University of Žilina · ObsahZoznam01112131415161718191 Zoznamintegrálov–príklady101–200...
Transcript of University of Žilina · ObsahZoznam01112131415161718191 Zoznamintegrálov–príklady101–200...
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Obsah Zoznam 01 11 21 31 41 51 61 71 81 91
Matematická analýza 1
2018/2019
13. Neurčitý integrálRiešené príklady
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Obsah Zoznam 01 11 21 31 41 51 61 71 81 91
Obsah – príklady 101–200
1 Riešené príklady 101–1102 Riešené príklady 111–1203 Riešené príklady 121–1304 Riešené príklady 131–1405 Riešené príklady 141–1506 Riešené príklady 151–1607 Riešené príklady 161–1708 Riešené príklady 171–1809 Riešené príklady 181–19010 Riešené príklady 191–200
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mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
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Obsah Zoznam 01 11 21 31 41 51 61 71 81 91
Zoznam integrálov – príklady 101–200101.
∫dx
x2+4x+5 102.
∫dx
x2+4x+3 103.
∫dx
x2−4x+6 104.
∫dx
x2−4x+2 105.
∫dx
x2+4x+4 106.
∫dx√
x2+4x+4107.
∫dx√
x2+4x+5108.
∫dx√
x2+4x+3
109.
∫dx√
x2−4x+6110.
∫dx√
x2−4x+2111.
∫dx√
−x2+4x−5112.
∫dx√
−x2+4x−3113.
∫dx√
x(x−1)114.
∫dx√
x(2−x)115.
∫√
x2+4x +5 dx
116.
∫√
x2+4x +3 dx 117.∫√
x2−4x +6 dx 118.∫√
x2−4x +2 dx 119.∫√−x2+4x−3 dx 120.
∫ √x(2−x) dx 121.
∫dx
(x2+a2)n 122.
∫dx
(x2+a2)2
123.
∫dx
(x2+a2)3 124.
∫dx
(x2+a2)4 125.
∫dx
(x2−a2)n 126.
∫dx
(x2−a2)2 127.
∫dx
(x2−a2)3 128.
∫dx
(x2−a2)4 129.
∫dx
(x2+4x+5)2 130.
∫dx
(x2+4x+3)2 131.
∫x2 dx
(x2+a2)2
132.
∫x2 dx
(x2−a2)2 133.
∫x dx
x2+a2 134.
∫x dx
(x2+a2)n 135.
∫x dx
x2−a2 136.
∫x dx
(x2−a2)n 137.
∫xn dxx−1 138.
∫x dxx−1 139.
∫x2 dxx−1 140.
∫x9 dxx−1 141.
∫dx
(1−x)x2
142.
∫dx
x6(1+x2) 143.
∫(x−2)4 dx
(x−1)2 144.
∫(x−1)4 dx
(x−2)2 145.
∫dx
x3−7x−6 146.
∫dx
x3−2x2−x+2 147.
∫dx
x3−3x−2 148.
∫dx
x3+x2−x−1 149.
∫dx
x3−3x2+4x−2
150.
∫dx
x3−3x2+3x−1 151.
∫dx
x3−2x−4 152.
∫dx
x6+1 153.
∫x dxx6+1 154.
∫x2 dxx6+1 155.
∫x3 dxx6+1 156.
∫x4 dxx6+1 157.
∫x5 dxx6+1 158.
∫x6 dxx6+1 159.
∫dx
2x +1 160.
∫dx√2x +1
161.
∫(1+x) dx√
1−x2162.
∫ √1+x1−x dx 163.
∫ √1−x1+x dx 164.
∫ √x+1x−1 dx 165.
∫ √x−1x+1 dx 166.
∫ √(1+x1−x
)3 dx 167.∫ √( 1−x1+x )3 dx168.
∫ √(x+1x−1
)3 dx 169.∫ √( x−1x+1 )3 dx 170.∫ √1−√x√1+√x dx 171.∫ dx√x+1+ 3√x+1 172.∫ arcsin√ 1+x1−x dx 173.∫ arcsin√ 1−x1+x dx174.
∫arcsin
√x+1x−1 dx 175.
∫arcsin
√x−1x+1 dx 176.
∫dx√
x−3+√
x−5 177.
∫dx√
x−3−√
x−5 178.
∫dx√
x−3+√5−x 179.
∫1+√
1−x2
1−√
1−x2dx
180.
∫ √ x1−x√
x dx 181.∫
dx3√x+ 4√
x 182.
∫dx
6√x+ 4√
x 183.
∫dx
3√x+ 5√
x 184.
∫dx
3√x+1 185.
∫ [√x3 − 1√x
]dx 186.
∫x−1
(√
x+ 3√
x2)xdx 187.
∫dx
(x−1)√
x−2
188.
∫dx
(x+1)√1−x 189.
∫(x−1)
√x−2 dx 190.
∫√1−x
x+1 dx 191.∫
dx(x−√
x2−1)2 192.∫ x5 dx√x2+1 193.
∫x5 dx√
x3+1194.
∫x5 dx√x2−1
195.
∫x5 dx√x3−1
196.
∫x5 dx√1−x2
197.
∫x5 dx√1−x3
198.
∫ √1−x2x2 dx 199.
∫dx
x√
x4+x2+1200.
∫dx
x√
x6+x3+1
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
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Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 101, 102∫dx
x2+4x+5 = arctg (x +2) + c
=∫
dxx2+4x+4+1 =
∫dx
(x+2)2+1 =[Subst. t =x +2 x ∈R
dt =dx t∈R
]
=∫
dtt2+1 =arctg t + c = arctg (x +2) + c, x ∈R, c∈R.
∫dx
x2+4x+3 =12 ln∣∣ x+1
x+3∣∣+ c
=∫
dxx2+4x+4−1 =
∫dx
(x+2)2−1 =[ Subst. t =x +2 x ∈R−{−1,− 3}
dt =dx t∈R−{±1}
]
=∫
dtt2−1 =
12 ln∣∣ t−1
t+1∣∣+ c = 12 ln ∣∣ x+2−1x+2+1 ∣∣+ c = 12 ln ∣∣ x+1x+3 ∣∣+ c,
x ∈R−{−1,− 3}, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
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Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 101, 102∫dx
x2+4x+5 = arctg (x +2) + c
=∫
dxx2+4x+4+1 =
∫dx
(x+2)2+1 =[Subst. t =x +2 x ∈R
dt =dx t∈R
]
=∫
dtt2+1 =arctg t + c = arctg (x +2) + c, x ∈R, c∈R.
∫dx
x2+4x+3 =12 ln∣∣ x+1
x+3∣∣+ c
=∫
dxx2+4x+4−1 =
∫dx
(x+2)2−1 =[ Subst. t =x +2 x ∈R−{−1,− 3}
dt =dx t∈R−{±1}
]
=∫
dtt2−1 =
12 ln∣∣ t−1
t+1∣∣+ c = 12 ln ∣∣ x+2−1x+2+1 ∣∣+ c = 12 ln ∣∣ x+1x+3 ∣∣+ c,
x ∈R−{−1,− 3}, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
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Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 101, 102∫dx
x2+4x+5 = arctg (x +2) + c
=∫
dxx2+4x+4+1 =
∫dx
(x+2)2+1 =[Subst. t =x +2 x ∈R
dt =dx t∈R
]
=∫
dtt2+1 =arctg t + c = arctg (x +2) + c, x ∈R, c∈R.
∫dx
x2+4x+3 =12 ln∣∣ x+1
x+3∣∣+ c
=∫
dxx2+4x+4−1 =
∫dx
(x+2)2−1 =[ Subst. t =x +2 x ∈R−{−1,− 3}
dt =dx t∈R−{±1}
]
=∫
dtt2−1 =
12 ln∣∣ t−1
t+1∣∣+ c = 12 ln ∣∣ x+2−1x+2+1 ∣∣+ c = 12 ln ∣∣ x+1x+3 ∣∣+ c,
x ∈R−{−1,− 3}, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
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Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 101, 102∫dx
x2+4x+5 = arctg (x +2) + c
=∫
dxx2+4x+4+1 =
∫dx
(x+2)2+1 =[Subst. t =x +2 x ∈R
dt =dx t∈R
]
=∫
dtt2+1 =arctg t + c = arctg (x +2) + c, x ∈R, c∈R.
∫dx
x2+4x+3 =12 ln∣∣ x+1
x+3∣∣+ c
=∫
dxx2+4x+4−1 =
∫dx
(x+2)2−1 =[ Subst. t =x +2 x ∈R−{−1,− 3}
dt =dx t∈R−{±1}
]
=∫
dtt2−1 =
12 ln∣∣ t−1
t+1∣∣+ c = 12 ln ∣∣ x+2−1x+2+1 ∣∣+ c = 12 ln ∣∣ x+1x+3 ∣∣+ c,
x ∈R−{−1,− 3}, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
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Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 101, 102∫dx
x2+4x+5 = arctg (x +2) + c
=∫
dxx2+4x+4+1 =
∫dx
(x+2)2+1 =[Subst. t =x +2 x ∈R
dt =dx t∈R
]
=∫
dtt2+1 =arctg t + c = arctg (x +2) + c, x ∈R, c∈R.
∫dx
x2+4x+3 =12 ln∣∣ x+1
x+3∣∣+ c
=∫
dxx2+4x+4−1 =
∫dx
(x+2)2−1 =[ Subst. t =x +2 x ∈R−{−1,− 3}
dt =dx t∈R−{±1}
]
=∫
dtt2−1 =
12 ln∣∣ t−1
t+1∣∣+ c = 12 ln ∣∣ x+2−1x+2+1 ∣∣+ c = 12 ln ∣∣ x+1x+3 ∣∣+ c,
x ∈R−{−1,− 3}, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 101, 102∫dx
x2+4x+5 = arctg (x +2) + c
=∫
dxx2+4x+4+1 =
∫dx
(x+2)2+1 =[Subst. t =x +2 x ∈R
dt =dx t∈R
]
=∫
dtt2+1 =arctg t + c = arctg (x +2) + c, x ∈R, c∈R.
∫dx
x2+4x+3 =12 ln∣∣ x+1
x+3∣∣+ c
=∫
dxx2+4x+4−1 =
∫dx
(x+2)2−1 =[ Subst. t =x +2 x ∈R−{−1,− 3}
dt =dx t∈R−{±1}
]
=∫
dtt2−1 =
12 ln∣∣ t−1
t+1∣∣+ c = 12 ln ∣∣ x+2−1x+2+1 ∣∣+ c = 12 ln ∣∣ x+1x+3 ∣∣+ c,
x ∈R−{−1,− 3}, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 101, 102∫dx
x2+4x+5 = arctg (x +2) + c
=∫
dxx2+4x+4+1 =
∫dx
(x+2)2+1 =[Subst. t =x +2 x ∈R
dt =dx t∈R
]
=∫
dtt2+1 =arctg t + c = arctg (x +2) + c, x ∈R, c∈R.
∫dx
x2+4x+3 =12 ln∣∣ x+1
x+3∣∣+ c
=∫
dxx2+4x+4−1 =
∫dx
(x+2)2−1 =[ Subst. t =x +2 x ∈R−{−1,− 3}
dt =dx t∈R−{±1}
]
=∫
dtt2−1 =
12 ln∣∣ t−1
t+1∣∣+ c = 12 ln ∣∣ x+2−1x+2+1 ∣∣+ c = 12 ln ∣∣ x+1x+3 ∣∣+ c,
x ∈R−{−1,− 3}, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
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Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 103, 104∫dx
x2−4x+6 = arctg (x +2) + c
=∫
dxx2−4x+4+2 =
∫dx
(x−2)2+(√2)2 =
[ Subst. t =x−2 x ∈Rdt =dx t∈R
]
=∫
dtt2+(√2)2 =
1√2 arctg
t√2 + c =
1√2 arctg
x−2√2 + c, x ∈R, c∈R.
∫dx
x2−4x+2 =12 ln∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c
=∫
dxx2−4x+4−2 =
∫dx
(x−2)2−(√2)2 =
[Subst. t =x−2 x ∈R−{2±
√2}
dt =dx t∈R−{±√2}
]
=∫
dtt2−(√2)2 =
12√2 ln∣∣∣ t−√2t+√2 ∣∣∣+ c = 12√2 ln ∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c,
x ∈R−{2±√2}, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
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Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 103, 104∫dx
x2−4x+6 = arctg (x +2) + c
=∫
dxx2−4x+4+2 =
∫dx
(x−2)2+(√2)2 =
[ Subst. t =x−2 x ∈Rdt =dx t∈R
]
=∫
dtt2+(√2)2 =
1√2 arctg
t√2 + c =
1√2 arctg
x−2√2 + c, x ∈R, c∈R.
∫dx
x2−4x+2 =12 ln∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c
=∫
dxx2−4x+4−2 =
∫dx
(x−2)2−(√2)2 =
[Subst. t =x−2 x ∈R−{2±
√2}
dt =dx t∈R−{±√2}
]
=∫
dtt2−(√2)2 =
12√2 ln∣∣∣ t−√2t+√2 ∣∣∣+ c = 12√2 ln ∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c,
x ∈R−{2±√2}, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 103, 104∫dx
x2−4x+6 = arctg (x +2) + c
=∫
dxx2−4x+4+2 =
∫dx
(x−2)2+(√2)2 =
[ Subst. t =x−2 x ∈Rdt =dx t∈R
]
=∫
dtt2+(√2)2 =
1√2 arctg
t√2 + c =
1√2 arctg
x−2√2 + c, x ∈R, c∈R.
∫dx
x2−4x+2 =12 ln∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c
=∫
dxx2−4x+4−2 =
∫dx
(x−2)2−(√2)2 =
[Subst. t =x−2 x ∈R−{2±
√2}
dt =dx t∈R−{±√2}
]
=∫
dtt2−(√2)2 =
12√2 ln∣∣∣ t−√2t+√2 ∣∣∣+ c = 12√2 ln ∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c,
x ∈R−{2±√2}, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 103, 104∫dx
x2−4x+6 = arctg (x +2) + c
=∫
dxx2−4x+4+2 =
∫dx
(x−2)2+(√2)2 =
[ Subst. t =x−2 x ∈Rdt =dx t∈R
]
=∫
dtt2+(√2)2 =
1√2 arctg
t√2 + c =
1√2 arctg
x−2√2 + c, x ∈R, c∈R.
∫dx
x2−4x+2 =12 ln∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c
=∫
dxx2−4x+4−2 =
∫dx
(x−2)2−(√2)2 =
[Subst. t =x−2 x ∈R−{2±
√2}
dt =dx t∈R−{±√2}
]
=∫
dtt2−(√2)2 =
12√2 ln∣∣∣ t−√2t+√2 ∣∣∣+ c = 12√2 ln ∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c,
x ∈R−{2±√2}, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 103, 104∫dx
x2−4x+6 = arctg (x +2) + c
=∫
dxx2−4x+4+2 =
∫dx
(x−2)2+(√2)2 =
[ Subst. t =x−2 x ∈Rdt =dx t∈R
]
=∫
dtt2+(√2)2 =
1√2 arctg
t√2 + c =
1√2 arctg
x−2√2 + c, x ∈R, c∈R.
∫dx
x2−4x+2 =12 ln∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c
=∫
dxx2−4x+4−2 =
∫dx
(x−2)2−(√2)2 =
[Subst. t =x−2 x ∈R−{2±
√2}
dt =dx t∈R−{±√2}
]
=∫
dtt2−(√2)2 =
12√2 ln∣∣∣ t−√2t+√2 ∣∣∣+ c = 12√2 ln ∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c,
x ∈R−{2±√2}, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 103, 104∫dx
x2−4x+6 = arctg (x +2) + c
=∫
dxx2−4x+4+2 =
∫dx
(x−2)2+(√2)2 =
[ Subst. t =x−2 x ∈Rdt =dx t∈R
]
=∫
dtt2+(√2)2 =
1√2 arctg
t√2 + c =
1√2 arctg
x−2√2 + c, x ∈R, c∈R.
∫dx
x2−4x+2 =12 ln∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c
=∫
dxx2−4x+4−2 =
∫dx
(x−2)2−(√2)2 =
[Subst. t =x−2 x ∈R−{2±
√2}
dt =dx t∈R−{±√2}
]
=∫
dtt2−(√2)2 =
12√2 ln∣∣∣ t−√2t+√2 ∣∣∣+ c = 12√2 ln ∣∣∣ x−2−√2x−2+√2 ∣∣∣+ c,
x ∈R−{2±√2}, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
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Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 105, 106∫dx
x2+4x+4 = −1
x+2 + c
=∫
dx(x+2)2 =
[ Subst. t =x +2 x ∈R−{−2}dt =dx t∈R−{0}
]
=∫
dtt2 =
∫t−2 dt = t
−2+1
−2+1 +c =−1t +c = −
1x+2 +c, x ∈R−{−2}, c∈R.∫
dx√x2+4x+4 = sgn (x +2)·ln |x +2|+c
=∫
dx√(x+2)2
=∫
dx|x+2|=
[ Subst. t =x +2 x ∈R−{−2}dt =dx t∈R−{0}
]=∫
dt|t|
t>0 =∫
dtt = ln |t|+c = ln |x +2|+c, x ∈(−2;∞), c∈R,
t
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 105, 106∫dx
x2+4x+4 = −1
x+2 + c
=∫
dx(x+2)2 =
[ Subst. t =x +2 x ∈R−{−2}dt =dx t∈R−{0}
]
=∫
dtt2 =
∫t−2 dt = t
−2+1
−2+1 +c =−1t +c = −
1x+2 +c, x ∈R−{−2}, c∈R.∫
dx√x2+4x+4 = sgn (x +2)·ln |x +2|+c
=∫
dx√(x+2)2
=∫
dx|x+2|=
[ Subst. t =x +2 x ∈R−{−2}dt =dx t∈R−{0}
]=∫
dt|t|
t>0 =∫
dtt = ln |t|+c = ln |x +2|+c, x ∈(−2;∞), c∈R,
t
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 105, 106∫dx
x2+4x+4 = −1
x+2 + c
=∫
dx(x+2)2 =
[ Subst. t =x +2 x ∈R−{−2}dt =dx t∈R−{0}
]
=∫
dtt2 =
∫t−2 dt = t
−2+1
−2+1 +c =−1t +c = −
1x+2 +c, x ∈R−{−2}, c∈R.∫
dx√x2+4x+4 = sgn (x +2)·ln |x +2|+c
=∫
dx√(x+2)2
=∫
dx|x+2|=
[ Subst. t =x +2 x ∈R−{−2}dt =dx t∈R−{0}
]=∫
dt|t|
t>0 =∫
dtt = ln |t|+c = ln |x +2|+c, x ∈(−2;∞), c∈R,
t
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 105, 106∫dx
x2+4x+4 = −1
x+2 + c
=∫
dx(x+2)2 =
[ Subst. t =x +2 x ∈R−{−2}dt =dx t∈R−{0}
]
=∫
dtt2 =
∫t−2 dt = t
−2+1
−2+1 +c =−1t +c = −
1x+2 +c, x ∈R−{−2}, c∈R.∫
dx√x2+4x+4 = sgn (x +2)·ln |x +2|+c
=∫
dx√(x+2)2
=∫
dx|x+2|=
[ Subst. t =x +2 x ∈R−{−2}dt =dx t∈R−{0}
]=∫
dt|t|
t>0 =∫
dtt = ln |t|+c = ln |x +2|+c, x ∈(−2;∞), c∈R,
t
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 105, 106∫dx
x2+4x+4 = −1
x+2 + c
=∫
dx(x+2)2 =
[ Subst. t =x +2 x ∈R−{−2}dt =dx t∈R−{0}
]
=∫
dtt2 =
∫t−2 dt = t
−2+1
−2+1 +c =−1t +c = −
1x+2 +c, x ∈R−{−2}, c∈R.∫
dx√x2+4x+4 = sgn (x +2)·ln |x +2|+c
=∫
dx√(x+2)2
=∫
dx|x+2|=
[ Subst. t =x +2 x ∈R−{−2}dt =dx t∈R−{0}
]=∫
dt|t|
t>0 =∫
dtt = ln |t|+c = ln |x +2|+c, x ∈(−2;∞), c∈R,
t
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 105, 106∫dx
x2+4x+4 = −1
x+2 + c
=∫
dx(x+2)2 =
[ Subst. t =x +2 x ∈R−{−2}dt =dx t∈R−{0}
]
=∫
dtt2 =
∫t−2 dt = t
−2+1
−2+1 +c =−1t +c = −
1x+2 +c, x ∈R−{−2}, c∈R.∫
dx√x2+4x+4 = sgn (x +2)·ln |x +2|+c
=∫
dx√(x+2)2
=∫
dx|x+2|=
[ Subst. t =x +2 x ∈R−{−2}dt =dx t∈R−{0}
]=∫
dt|t|
t>0 =∫
dtt = ln |t|+c = ln |x +2|+c, x ∈(−2;∞), c∈R,
t
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 105, 106∫dx
x2+4x+4 = −1
x+2 + c
=∫
dx(x+2)2 =
[ Subst. t =x +2 x ∈R−{−2}dt =dx t∈R−{0}
]
=∫
dtt2 =
∫t−2 dt = t
−2+1
−2+1 +c =−1t +c = −
1x+2 +c, x ∈R−{−2}, c∈R.∫
dx√x2+4x+4 = sgn (x +2)·ln |x +2|+c
=∫
dx√(x+2)2
=∫
dx|x+2|=
[ Subst. t =x +2 x ∈R−{−2}dt =dx t∈R−{0}
]=∫
dt|t|
t>0 =∫
dtt = ln |t|+c = ln |x +2|+c, x ∈(−2;∞), c∈R,
t
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 107, 108∫dx√
x2+4x+5 = ln (x +2+√
x2+4x +5)+c
=∫
dx√x2+4x+4+1 =
∫dx√
(x+2)2+1=[Subst. t =x +2 x ∈R
dt =dx t∈R
]
=∫
dt√t2+1 = ln (t+
√t2+1)+c = ln (x +2+
√(x +2)2+1)+c
= ln (x +2+√
x2+4x +5)+c, x ∈R, c∈R.∫dx√
x2+4x+3 = ln |x +2+√
x2+4x +3|+c
=∫
dx√x2+4x+4−1
=∫
dx√(x+2)2−1
=[ Subst. t =x +2 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx x ∈(−1;∞) , t∈(1;∞)
]
=∫
dt√t2−1
= ln |t+√
t2−1|+c = ln |x +2+√
(x +2)2−1|+c
= ln |x +2+√
x2+4x +3|+c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 107, 108∫dx√
x2+4x+5 = ln (x +2+√
x2+4x +5)+c
=∫
dx√x2+4x+4+1 =
∫dx√
(x+2)2+1=[Subst. t =x +2 x ∈R
dt =dx t∈R
]
=∫
dt√t2+1 = ln (t+
√t2+1)+c = ln (x +2+
√(x +2)2+1)+c
= ln (x +2+√
x2+4x +5)+c, x ∈R, c∈R.∫dx√
x2+4x+3 = ln |x +2+√
x2+4x +3|+c
=∫
dx√x2+4x+4−1
=∫
dx√(x+2)2−1
=[ Subst. t =x +2 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx x ∈(−1;∞) , t∈(1;∞)
]
=∫
dt√t2−1
= ln |t+√
t2−1|+c = ln |x +2+√
(x +2)2−1|+c
= ln |x +2+√
x2+4x +3|+c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 107, 108∫dx√
x2+4x+5 = ln (x +2+√
x2+4x +5)+c
=∫
dx√x2+4x+4+1 =
∫dx√
(x+2)2+1=[Subst. t =x +2 x ∈R
dt =dx t∈R
]
=∫
dt√t2+1 = ln (t+
√t2+1)+c = ln (x +2+
√(x +2)2+1)+c
= ln (x +2+√
x2+4x +5)+c, x ∈R, c∈R.∫dx√
x2+4x+3 = ln |x +2+√
x2+4x +3|+c
=∫
dx√x2+4x+4−1
=∫
dx√(x+2)2−1
=[ Subst. t =x +2 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx x ∈(−1;∞) , t∈(1;∞)
]
=∫
dt√t2−1
= ln |t+√
t2−1|+c = ln |x +2+√
(x +2)2−1|+c
= ln |x +2+√
x2+4x +3|+c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 107, 108∫dx√
x2+4x+5 = ln (x +2+√
x2+4x +5)+c
=∫
dx√x2+4x+4+1 =
∫dx√
(x+2)2+1=[Subst. t =x +2 x ∈R
dt =dx t∈R
]
=∫
dt√t2+1 = ln (t+
√t2+1)+c = ln (x +2+
√(x +2)2+1)+c
= ln (x +2+√
x2+4x +5)+c, x ∈R, c∈R.∫dx√
x2+4x+3 = ln |x +2+√
x2+4x +3|+c
=∫
dx√x2+4x+4−1
=∫
dx√(x+2)2−1
=[ Subst. t =x +2 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx x ∈(−1;∞) , t∈(1;∞)
]
=∫
dt√t2−1
= ln |t+√
t2−1|+c = ln |x +2+√
(x +2)2−1|+c
= ln |x +2+√
x2+4x +3|+c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 107, 108∫dx√
x2+4x+5 = ln (x +2+√
x2+4x +5)+c
=∫
dx√x2+4x+4+1 =
∫dx√
(x+2)2+1=[Subst. t =x +2 x ∈R
dt =dx t∈R
]
=∫
dt√t2+1 = ln (t+
√t2+1)+c = ln (x +2+
√(x +2)2+1)+c
= ln (x +2+√
x2+4x +5)+c, x ∈R, c∈R.∫dx√
x2+4x+3 = ln |x +2+√
x2+4x +3|+c
=∫
dx√x2+4x+4−1
=∫
dx√(x+2)2−1
=[ Subst. t =x +2 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx x ∈(−1;∞) , t∈(1;∞)
]
=∫
dt√t2−1
= ln |t+√
t2−1|+c = ln |x +2+√
(x +2)2−1|+c
= ln |x +2+√
x2+4x +3|+c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 107, 108∫dx√
x2+4x+5 = ln (x +2+√
x2+4x +5)+c
=∫
dx√x2+4x+4+1 =
∫dx√
(x+2)2+1=[Subst. t =x +2 x ∈R
dt =dx t∈R
]
=∫
dt√t2+1 = ln (t+
√t2+1)+c = ln (x +2+
√(x +2)2+1)+c
= ln (x +2+√
x2+4x +5)+c, x ∈R, c∈R.∫dx√
x2+4x+3 = ln |x +2+√
x2+4x +3|+c
=∫
dx√x2+4x+4−1
=∫
dx√(x+2)2−1
=[ Subst. t =x +2 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx x ∈(−1;∞) , t∈(1;∞)
]
=∫
dt√t2−1
= ln |t+√
t2−1|+c = ln |x +2+√
(x +2)2−1|+c
= ln |x +2+√
x2+4x +3|+c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 107, 108∫dx√
x2+4x+5 = ln (x +2+√
x2+4x +5)+c
=∫
dx√x2+4x+4+1 =
∫dx√
(x+2)2+1=[Subst. t =x +2 x ∈R
dt =dx t∈R
]
=∫
dt√t2+1 = ln (t+
√t2+1)+c = ln (x +2+
√(x +2)2+1)+c
= ln (x +2+√
x2+4x +5)+c, x ∈R, c∈R.∫dx√
x2+4x+3 = ln |x +2+√
x2+4x +3|+c
=∫
dx√x2+4x+4−1
=∫
dx√(x+2)2−1
=[ Subst. t =x +2 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx x ∈(−1;∞) , t∈(1;∞)
]
=∫
dt√t2−1
= ln |t+√
t2−1|+c = ln |x +2+√
(x +2)2−1|+c
= ln |x +2+√
x2+4x +3|+c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 109, 110∫dx√
x2−4x+6= ln (x−2+
√x2−4x +6)+c
=∫
dx√x2−4x+4+2
=∫
dx√(x−2)2+(
√2)2
=[Subst. t =x−2 x ∈R
dt =dx t∈R
]
=∫
dt√t2+(√2)2
= ln(t+√
t2+(√2)2)+c = ln (x +2+
√(x−2)2+4)+c
= ln (x−2+√
x2−4x +6)+c, x ∈R, c∈R.∫dx√
x2−4x+2= ln |x−2+
√x2−4x +2|+c
=∫
dx√x2−4x+4−2
=∫
dx√(x−2)2−(
√2)2
=[Subst. t =x−2 x ∈(−∞; 2−
√2), t∈(−∞;−
√2)
dt =dx x ∈(2+√2;∞), t∈(
√2;∞)
]
=∫
dt√t2−(√2)2
= ln∣∣t+√t2−(√2)2∣∣+c = ln |x−2+√(x +2)2−2|+c
= ln |x−2+√
x2−4x +2|+c, x ∈(−∞; 2−√2) ∪ (2+
√2;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 109, 110∫dx√
x2−4x+6= ln (x−2+
√x2−4x +6)+c
=∫
dx√x2−4x+4+2
=∫
dx√(x−2)2+(
√2)2
=[Subst. t =x−2 x ∈R
dt =dx t∈R
]
=∫
dt√t2+(√2)2
= ln(t+√
t2+(√2)2)+c = ln (x +2+
√(x−2)2+4)+c
= ln (x−2+√
x2−4x +6)+c, x ∈R, c∈R.∫dx√
x2−4x+2= ln |x−2+
√x2−4x +2|+c
=∫
dx√x2−4x+4−2
=∫
dx√(x−2)2−(
√2)2
=[Subst. t =x−2 x ∈(−∞; 2−
√2), t∈(−∞;−
√2)
dt =dx x ∈(2+√2;∞), t∈(
√2;∞)
]
=∫
dt√t2−(√2)2
= ln∣∣t+√t2−(√2)2∣∣+c = ln |x−2+√(x +2)2−2|+c
= ln |x−2+√
x2−4x +2|+c, x ∈(−∞; 2−√2) ∪ (2+
√2;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 109, 110∫dx√
x2−4x+6= ln (x−2+
√x2−4x +6)+c
=∫
dx√x2−4x+4+2
=∫
dx√(x−2)2+(
√2)2
=[Subst. t =x−2 x ∈R
dt =dx t∈R
]
=∫
dt√t2+(√2)2
= ln(t+√
t2+(√2)2)+c = ln (x +2+
√(x−2)2+4)+c
= ln (x−2+√
x2−4x +6)+c, x ∈R, c∈R.∫dx√
x2−4x+2= ln |x−2+
√x2−4x +2|+c
=∫
dx√x2−4x+4−2
=∫
dx√(x−2)2−(
√2)2
=[Subst. t =x−2 x ∈(−∞; 2−
√2), t∈(−∞;−
√2)
dt =dx x ∈(2+√2;∞), t∈(
√2;∞)
]
=∫
dt√t2−(√2)2
= ln∣∣t+√t2−(√2)2∣∣+c = ln |x−2+√(x +2)2−2|+c
= ln |x−2+√
x2−4x +2|+c, x ∈(−∞; 2−√2) ∪ (2+
√2;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 109, 110∫dx√
x2−4x+6= ln (x−2+
√x2−4x +6)+c
=∫
dx√x2−4x+4+2
=∫
dx√(x−2)2+(
√2)2
=[Subst. t =x−2 x ∈R
dt =dx t∈R
]
=∫
dt√t2+(√2)2
= ln(t+√
t2+(√2)2)+c = ln (x +2+
√(x−2)2+4)+c
= ln (x−2+√
x2−4x +6)+c, x ∈R, c∈R.∫dx√
x2−4x+2= ln |x−2+
√x2−4x +2|+c
=∫
dx√x2−4x+4−2
=∫
dx√(x−2)2−(
√2)2
=[Subst. t =x−2 x ∈(−∞; 2−
√2), t∈(−∞;−
√2)
dt =dx x ∈(2+√2;∞), t∈(
√2;∞)
]
=∫
dt√t2−(√2)2
= ln∣∣t+√t2−(√2)2∣∣+c = ln |x−2+√(x +2)2−2|+c
= ln |x−2+√
x2−4x +2|+c, x ∈(−∞; 2−√2) ∪ (2+
√2;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 109, 110∫dx√
x2−4x+6= ln (x−2+
√x2−4x +6)+c
=∫
dx√x2−4x+4+2
=∫
dx√(x−2)2+(
√2)2
=[Subst. t =x−2 x ∈R
dt =dx t∈R
]
=∫
dt√t2+(√2)2
= ln(t+√
t2+(√2)2)+c = ln (x +2+
√(x−2)2+4)+c
= ln (x−2+√
x2−4x +6)+c, x ∈R, c∈R.∫dx√
x2−4x+2= ln |x−2+
√x2−4x +2|+c
=∫
dx√x2−4x+4−2
=∫
dx√(x−2)2−(
√2)2
=[Subst. t =x−2 x ∈(−∞; 2−
√2), t∈(−∞;−
√2)
dt =dx x ∈(2+√2;∞), t∈(
√2;∞)
]
=∫
dt√t2−(√2)2
= ln∣∣t+√t2−(√2)2∣∣+c = ln |x−2+√(x +2)2−2|+c
= ln |x−2+√
x2−4x +2|+c, x ∈(−∞; 2−√2) ∪ (2+
√2;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 109, 110∫dx√
x2−4x+6= ln (x−2+
√x2−4x +6)+c
=∫
dx√x2−4x+4+2
=∫
dx√(x−2)2+(
√2)2
=[Subst. t =x−2 x ∈R
dt =dx t∈R
]
=∫
dt√t2+(√2)2
= ln(t+√
t2+(√2)2)+c = ln (x +2+
√(x−2)2+4)+c
= ln (x−2+√
x2−4x +6)+c, x ∈R, c∈R.∫dx√
x2−4x+2= ln |x−2+
√x2−4x +2|+c
=∫
dx√x2−4x+4−2
=∫
dx√(x−2)2−(
√2)2
=[Subst. t =x−2 x ∈(−∞; 2−
√2), t∈(−∞;−
√2)
dt =dx x ∈(2+√2;∞), t∈(
√2;∞)
]
=∫
dt√t2−(√2)2
= ln∣∣t+√t2−(√2)2∣∣+c = ln |x−2+√(x +2)2−2|+c
= ln |x−2+√
x2−4x +2|+c, x ∈(−∞; 2−√2) ∪ (2+
√2;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 01–02 03–04 05–06 07–08 09–10
Riešené príklady – 109, 110∫dx√
x2−4x+6= ln (x−2+
√x2−4x +6)+c
=∫
dx√x2−4x+4+2
=∫
dx√(x−2)2+(
√2)2
=[Subst. t =x−2 x ∈R
dt =dx t∈R
]
=∫
dt√t2+(√2)2
= ln(t+√
t2+(√2)2)+c = ln (x +2+
√(x−2)2+4)+c
= ln (x−2+√
x2−4x +6)+c, x ∈R, c∈R.∫dx√
x2−4x+2= ln |x−2+
√x2−4x +2|+c
=∫
dx√x2−4x+4−2
=∫
dx√(x−2)2−(
√2)2
=[Subst. t =x−2 x ∈(−∞; 2−
√2), t∈(−∞;−
√2)
dt =dx x ∈(2+√2;∞), t∈(
√2;∞)
]
=∫
dt√t2−(√2)2
= ln∣∣t+√t2−(√2)2∣∣+c = ln |x−2+√(x +2)2−2|+c
= ln |x−2+√
x2−4x +2|+c, x ∈(−∞; 2−√2) ∪ (2+
√2;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 111, 112∫dx√
−x2+4x−5nemá riešenie
=∫
dx√−(x2−4x+5)
=∫
dx√−(x2−4x+4+1)
=∫
dx√−(x−2)2−1
=[−(x−2)2−1≥1>0
pre všetky x ∈R
].
Neexistuje riešenie pre žiadne x ∈R.∫dx√
−x2+4x−3= arcsin (x−2)+c1 = − arccos (x−2)+c2
=∫
dx√−(x2−4x+3)
=∫
dx√−(x2−4x+4−1)
=∫
dx√−(x−2)2+1
=∫
dx√1−(x−2)2
=[ Subst. t =x−2 x ∈(1; 3)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−2)+c1= − arccos (x−2)+c2, x ∈(1; 3), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 111, 112∫dx√
−x2+4x−5nemá riešenie
=∫
dx√−(x2−4x+5)
=∫
dx√−(x2−4x+4+1)
=∫
dx√−(x−2)2−1
=[−(x−2)2−1≥1>0
pre všetky x ∈R
].
Neexistuje riešenie pre žiadne x ∈R.∫dx√
−x2+4x−3= arcsin (x−2)+c1 = − arccos (x−2)+c2
=∫
dx√−(x2−4x+3)
=∫
dx√−(x2−4x+4−1)
=∫
dx√−(x−2)2+1
=∫
dx√1−(x−2)2
=[ Subst. t =x−2 x ∈(1; 3)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−2)+c1= − arccos (x−2)+c2, x ∈(1; 3), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 111, 112∫dx√
−x2+4x−5nemá riešenie
=∫
dx√−(x2−4x+5)
=∫
dx√−(x2−4x+4+1)
=∫
dx√−(x−2)2−1
=[−(x−2)2−1≥1>0
pre všetky x ∈R
].
Neexistuje riešenie pre žiadne x ∈R.∫dx√
−x2+4x−3= arcsin (x−2)+c1 = − arccos (x−2)+c2
=∫
dx√−(x2−4x+3)
=∫
dx√−(x2−4x+4−1)
=∫
dx√−(x−2)2+1
=∫
dx√1−(x−2)2
=[ Subst. t =x−2 x ∈(1; 3)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−2)+c1= − arccos (x−2)+c2, x ∈(1; 3), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 111, 112∫dx√
−x2+4x−5nemá riešenie
=∫
dx√−(x2−4x+5)
=∫
dx√−(x2−4x+4+1)
=∫
dx√−(x−2)2−1
=[−(x−2)2−1≥1>0
pre všetky x ∈R
].
Neexistuje riešenie pre žiadne x ∈R.∫dx√
−x2+4x−3= arcsin (x−2)+c1 = − arccos (x−2)+c2
=∫
dx√−(x2−4x+3)
=∫
dx√−(x2−4x+4−1)
=∫
dx√−(x−2)2+1
=∫
dx√1−(x−2)2
=[ Subst. t =x−2 x ∈(1; 3)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−2)+c1= − arccos (x−2)+c2, x ∈(1; 3), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 111, 112∫dx√
−x2+4x−5nemá riešenie
=∫
dx√−(x2−4x+5)
=∫
dx√−(x2−4x+4+1)
=∫
dx√−(x−2)2−1
=[−(x−2)2−1≥1>0
pre všetky x ∈R
].
Neexistuje riešenie pre žiadne x ∈R.∫dx√
−x2+4x−3= arcsin (x−2)+c1 = − arccos (x−2)+c2
=∫
dx√−(x2−4x+3)
=∫
dx√−(x2−4x+4−1)
=∫
dx√−(x−2)2+1
=∫
dx√1−(x−2)2
=[ Subst. t =x−2 x ∈(1; 3)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−2)+c1= − arccos (x−2)+c2, x ∈(1; 3), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 111, 112∫dx√
−x2+4x−5nemá riešenie
=∫
dx√−(x2−4x+5)
=∫
dx√−(x2−4x+4+1)
=∫
dx√−(x−2)2−1
=[−(x−2)2−1≥1>0
pre všetky x ∈R
].
Neexistuje riešenie pre žiadne x ∈R.∫dx√
−x2+4x−3= arcsin (x−2)+c1 = − arccos (x−2)+c2
=∫
dx√−(x2−4x+3)
=∫
dx√−(x2−4x+4−1)
=∫
dx√−(x−2)2+1
=∫
dx√1−(x−2)2
=[ Subst. t =x−2 x ∈(1; 3)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−2)+c1= − arccos (x−2)+c2, x ∈(1; 3), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 111, 112∫dx√
−x2+4x−5nemá riešenie
=∫
dx√−(x2−4x+5)
=∫
dx√−(x2−4x+4+1)
=∫
dx√−(x−2)2−1
=[−(x−2)2−1≥1>0
pre všetky x ∈R
].
Neexistuje riešenie pre žiadne x ∈R.∫dx√
−x2+4x−3= arcsin (x−2)+c1 = − arccos (x−2)+c2
=∫
dx√−(x2−4x+3)
=∫
dx√−(x2−4x+4−1)
=∫
dx√−(x−2)2+1
=∫
dx√1−(x−2)2
=[ Subst. t =x−2 x ∈(1; 3)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−2)+c1= − arccos (x−2)+c2, x ∈(1; 3), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 111, 112∫dx√
−x2+4x−5nemá riešenie
=∫
dx√−(x2−4x+5)
=∫
dx√−(x2−4x+4+1)
=∫
dx√−(x−2)2−1
=[−(x−2)2−1≥1>0
pre všetky x ∈R
].
Neexistuje riešenie pre žiadne x ∈R.∫dx√
−x2+4x−3= arcsin (x−2)+c1 = − arccos (x−2)+c2
=∫
dx√−(x2−4x+3)
=∫
dx√−(x2−4x+4−1)
=∫
dx√−(x−2)2+1
=∫
dx√1−(x−2)2
=[ Subst. t =x−2 x ∈(1; 3)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−2)+c1= − arccos (x−2)+c2, x ∈(1; 3), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 113, 114∫dx√
x(x−1)=∫
dx√x2−x
= ln |x− 12 +√
x2−x |+c1 = ln |2x−1+√
x2−x |+c2
=∫
dx√x2−2 x2 +
14−
14
=∫
dx√(x− 12 )2−(
12 )2
=[Subst. t =x− 12 x ∈(−∞; 0) , t∈
(−∞;− 12
)dt =dx x ∈(1;∞) , t∈
( 12 ;∞
) ]
=∫
dt√t2−( 12 )2
= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1
= ln∣∣∣ 2x−1+√x2−x2 ∣∣∣+c1= ln |2x−1+√x2−x |−ln 2+c1=[ c2 = c1−ln 2c1∈R, c2∈R ]
= ln |2x−1+√
x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√
x(2−x)=∫
dx√−x2+2x
= arcsin (x−1)+c1 = − arccos (x−1)+c2
=∫
dx√−(x2−2x+1)+1
=∫
dx√1−(x−1)2
=[ Subst. t =x−1 x ∈(0; 2)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−1)+c1= − arccos (x−1)+c2,x ∈(0; 2), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 113, 114∫dx√
x(x−1)=∫
dx√x2−x
= ln |x− 12 +√
x2−x |+c1 = ln |2x−1+√
x2−x |+c2
=∫
dx√x2−2 x2 +
14−
14
=∫
dx√(x− 12 )2−(
12 )2
=[Subst. t =x− 12 x ∈(−∞; 0) , t∈
(−∞;− 12
)dt =dx x ∈(1;∞) , t∈
( 12 ;∞
) ]
=∫
dt√t2−( 12 )2
= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1
= ln∣∣∣ 2x−1+√x2−x2 ∣∣∣+c1= ln |2x−1+√x2−x |−ln 2+c1=[ c2 = c1−ln 2c1∈R, c2∈R ]
= ln |2x−1+√
x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√
x(2−x)=∫
dx√−x2+2x
= arcsin (x−1)+c1 = − arccos (x−1)+c2
=∫
dx√−(x2−2x+1)+1
=∫
dx√1−(x−1)2
=[ Subst. t =x−1 x ∈(0; 2)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−1)+c1= − arccos (x−1)+c2,x ∈(0; 2), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 113, 114∫dx√
x(x−1)=∫
dx√x2−x
= ln |x− 12 +√
x2−x |+c1 = ln |2x−1+√
x2−x |+c2
=∫
dx√x2−2 x2 +
14−
14
=∫
dx√(x− 12 )2−(
12 )2
=[Subst. t =x− 12 x ∈(−∞; 0) , t∈
(−∞;− 12
)dt =dx x ∈(1;∞) , t∈
( 12 ;∞
) ]
=∫
dt√t2−( 12 )2
= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1
= ln∣∣∣ 2x−1+√x2−x2 ∣∣∣+c1= ln |2x−1+√x2−x |−ln 2+c1=[ c2 = c1−ln 2c1∈R, c2∈R ]
= ln |2x−1+√
x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√
x(2−x)=∫
dx√−x2+2x
= arcsin (x−1)+c1 = − arccos (x−1)+c2
=∫
dx√−(x2−2x+1)+1
=∫
dx√1−(x−1)2
=[ Subst. t =x−1 x ∈(0; 2)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−1)+c1= − arccos (x−1)+c2,x ∈(0; 2), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 113, 114∫dx√
x(x−1)=∫
dx√x2−x
= ln |x− 12 +√
x2−x |+c1 = ln |2x−1+√
x2−x |+c2
=∫
dx√x2−2 x2 +
14−
14
=∫
dx√(x− 12 )2−(
12 )2
=[Subst. t =x− 12 x ∈(−∞; 0) , t∈
(−∞;− 12
)dt =dx x ∈(1;∞) , t∈
( 12 ;∞
) ]
=∫
dt√t2−( 12 )2
= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1
= ln∣∣∣ 2x−1+√x2−x2 ∣∣∣+c1= ln |2x−1+√x2−x |−ln 2+c1=[ c2 = c1−ln 2c1∈R, c2∈R ]
= ln |2x−1+√
x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√
x(2−x)=∫
dx√−x2+2x
= arcsin (x−1)+c1 = − arccos (x−1)+c2
=∫
dx√−(x2−2x+1)+1
=∫
dx√1−(x−1)2
=[ Subst. t =x−1 x ∈(0; 2)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−1)+c1= − arccos (x−1)+c2,x ∈(0; 2), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 113, 114∫dx√
x(x−1)=∫
dx√x2−x
= ln |x− 12 +√
x2−x |+c1 = ln |2x−1+√
x2−x |+c2
=∫
dx√x2−2 x2 +
14−
14
=∫
dx√(x− 12 )2−(
12 )2
=[Subst. t =x− 12 x ∈(−∞; 0) , t∈
(−∞;− 12
)dt =dx x ∈(1;∞) , t∈
( 12 ;∞
) ]
=∫
dt√t2−( 12 )2
= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1
= ln∣∣∣ 2x−1+√x2−x2 ∣∣∣+c1= ln |2x−1+√x2−x |−ln 2+c1=[ c2 = c1−ln 2c1∈R, c2∈R ]
= ln |2x−1+√
x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√
x(2−x)=∫
dx√−x2+2x
= arcsin (x−1)+c1 = − arccos (x−1)+c2
=∫
dx√−(x2−2x+1)+1
=∫
dx√1−(x−1)2
=[ Subst. t =x−1 x ∈(0; 2)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−1)+c1= − arccos (x−1)+c2,x ∈(0; 2), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 113, 114∫dx√
x(x−1)=∫
dx√x2−x
= ln |x− 12 +√
x2−x |+c1 = ln |2x−1+√
x2−x |+c2
=∫
dx√x2−2 x2 +
14−
14
=∫
dx√(x− 12 )2−(
12 )2
=[Subst. t =x− 12 x ∈(−∞; 0) , t∈
(−∞;− 12
)dt =dx x ∈(1;∞) , t∈
( 12 ;∞
) ]
=∫
dt√t2−( 12 )2
= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1
= ln∣∣∣ 2x−1+√x2−x2 ∣∣∣+c1= ln |2x−1+√x2−x |−ln 2+c1=[ c2 = c1−ln 2c1∈R, c2∈R ]
= ln |2x−1+√
x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√
x(2−x)=∫
dx√−x2+2x
= arcsin (x−1)+c1 = − arccos (x−1)+c2
=∫
dx√−(x2−2x+1)+1
=∫
dx√1−(x−1)2
=[ Subst. t =x−1 x ∈(0; 2)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−1)+c1= − arccos (x−1)+c2,x ∈(0; 2), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 113, 114∫dx√
x(x−1)=∫
dx√x2−x
= ln |x− 12 +√
x2−x |+c1 = ln |2x−1+√
x2−x |+c2
=∫
dx√x2−2 x2 +
14−
14
=∫
dx√(x− 12 )2−(
12 )2
=[Subst. t =x− 12 x ∈(−∞; 0) , t∈
(−∞;− 12
)dt =dx x ∈(1;∞) , t∈
( 12 ;∞
) ]
=∫
dt√t2−( 12 )2
= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1
= ln∣∣∣ 2x−1+√x2−x2 ∣∣∣+c1= ln |2x−1+√x2−x |−ln 2+c1=[ c2 = c1−ln 2c1∈R, c2∈R ]
= ln |2x−1+√
x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√
x(2−x)=∫
dx√−x2+2x
= arcsin (x−1)+c1 = − arccos (x−1)+c2
=∫
dx√−(x2−2x+1)+1
=∫
dx√1−(x−1)2
=[ Subst. t =x−1 x ∈(0; 2)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−1)+c1= − arccos (x−1)+c2,x ∈(0; 2), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 113, 114∫dx√
x(x−1)=∫
dx√x2−x
= ln |x− 12 +√
x2−x |+c1 = ln |2x−1+√
x2−x |+c2
=∫
dx√x2−2 x2 +
14−
14
=∫
dx√(x− 12 )2−(
12 )2
=[Subst. t =x− 12 x ∈(−∞; 0) , t∈
(−∞;− 12
)dt =dx x ∈(1;∞) , t∈
( 12 ;∞
) ]
=∫
dt√t2−( 12 )2
= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1
= ln∣∣∣ 2x−1+√x2−x2 ∣∣∣+c1= ln |2x−1+√x2−x |−ln 2+c1=[ c2 = c1−ln 2c1∈R, c2∈R ]
= ln |2x−1+√
x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√
x(2−x)=∫
dx√−x2+2x
= arcsin (x−1)+c1 = − arccos (x−1)+c2
=∫
dx√−(x2−2x+1)+1
=∫
dx√1−(x−1)2
=[ Subst. t =x−1 x ∈(0; 2)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−1)+c1= − arccos (x−1)+c2,x ∈(0; 2), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 113, 114∫dx√
x(x−1)=∫
dx√x2−x
= ln |x− 12 +√
x2−x |+c1 = ln |2x−1+√
x2−x |+c2
=∫
dx√x2−2 x2 +
14−
14
=∫
dx√(x− 12 )2−(
12 )2
=[Subst. t =x− 12 x ∈(−∞; 0) , t∈
(−∞;− 12
)dt =dx x ∈(1;∞) , t∈
( 12 ;∞
) ]
=∫
dt√t2−( 12 )2
= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1
= ln∣∣∣ 2x−1+√x2−x2 ∣∣∣+c1= ln |2x−1+√x2−x |−ln 2+c1=[ c2 = c1−ln 2c1∈R, c2∈R ]
= ln |2x−1+√
x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√
x(2−x)=∫
dx√−x2+2x
= arcsin (x−1)+c1 = − arccos (x−1)+c2
=∫
dx√−(x2−2x+1)+1
=∫
dx√1−(x−1)2
=[ Subst. t =x−1 x ∈(0; 2)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−1)+c1= − arccos (x−1)+c2,x ∈(0; 2), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 113, 114∫dx√
x(x−1)=∫
dx√x2−x
= ln |x− 12 +√
x2−x |+c1 = ln |2x−1+√
x2−x |+c2
=∫
dx√x2−2 x2 +
14−
14
=∫
dx√(x− 12 )2−(
12 )2
=[Subst. t =x− 12 x ∈(−∞; 0) , t∈
(−∞;− 12
)dt =dx x ∈(1;∞) , t∈
( 12 ;∞
) ]
=∫
dt√t2−( 12 )2
= ln∣∣t+√t2−( 12 )2∣∣+c1= ln |x− 12 +√x2−x |+c1
= ln∣∣∣ 2x−1+√x2−x2 ∣∣∣+c1= ln |2x−1+√x2−x |−ln 2+c1=[ c2 = c1−ln 2c1∈R, c2∈R ]
= ln |2x−1+√
x2−x |+c2, x ∈(−∞; 0) ∪ (1;∞), c∈R.∫dx√
x(2−x)=∫
dx√−x2+2x
= arcsin (x−1)+c1 = − arccos (x−1)+c2
=∫
dx√−(x2−2x+1)+1
=∫
dx√1−(x−1)2
=[ Subst. t =x−1 x ∈(0; 2)
dt =dx t∈(−1; 1)
]=∫
dt√1−t2
=arcsin t+c1=− arccos t+c2= arcsin (x−1)+c1= − arccos (x−1)+c2,x ∈(0; 2), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 115, 116∫ √x2+4x +5 dx = (x+2)
√x2+4x+52 +
ln (x+2+√
x2+4x+5)2 +c
=[ Subst. t =x +2 x2+4x +5=x2+4x +4+1 x ∈R
dt =dx =(x +2)2+1= t2+1 t∈R
]=∫ √
t2+1 dt = t√
t2+12 +
12
∫dt√t2+1
= t√
t2+12 +
ln (t+√
t2+1)2 +c =
(x+2)√
(x+2)2+12 +
ln (x+2+√
(x+2)2+1)2 +c
= (x+2)√
x2+4x+52 +
ln (x+2+√
x2+4x+5)2 +c, x ∈R, c∈R.∫ √
x2+4x +3 dx = (x+2)√
x2+4x+32 −
ln |x+2+√
x2+4x+3|2 +c
=[ Subst. t =x +2 x2+4x +3=x2+4x +4−1 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx =(x +2)2−1= t2−1 x ∈(−1;∞) , t∈(1;∞)
]=∫ √
t2−1 dt
= t√
t2−12 −
12
∫dt√t2−1
= t√
t2−12 −
ln |t+√
t2−1|2 +c
= (x+2)√
x2+4x+32 −
ln |x+2+√
x2+4x+3|2 +c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 115, 116∫ √x2+4x +5 dx = (x+2)
√x2+4x+52 +
ln (x+2+√
x2+4x+5)2 +c
=[ Subst. t =x +2 x2+4x +5=x2+4x +4+1 x ∈R
dt =dx =(x +2)2+1= t2+1 t∈R
]=∫ √
t2+1 dt = t√
t2+12 +
12
∫dt√t2+1
= t√
t2+12 +
ln (t+√
t2+1)2 +c =
(x+2)√
(x+2)2+12 +
ln (x+2+√
(x+2)2+1)2 +c
= (x+2)√
x2+4x+52 +
ln (x+2+√
x2+4x+5)2 +c, x ∈R, c∈R.∫ √
x2+4x +3 dx = (x+2)√
x2+4x+32 −
ln |x+2+√
x2+4x+3|2 +c
=[ Subst. t =x +2 x2+4x +3=x2+4x +4−1 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx =(x +2)2−1= t2−1 x ∈(−1;∞) , t∈(1;∞)
]=∫ √
t2−1 dt
= t√
t2−12 −
12
∫dt√t2−1
= t√
t2−12 −
ln |t+√
t2−1|2 +c
= (x+2)√
x2+4x+32 −
ln |x+2+√
x2+4x+3|2 +c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 115, 116∫ √x2+4x +5 dx = (x+2)
√x2+4x+52 +
ln (x+2+√
x2+4x+5)2 +c
=[ Subst. t =x +2 x2+4x +5=x2+4x +4+1 x ∈R
dt =dx =(x +2)2+1= t2+1 t∈R
]=∫ √
t2+1 dt = t√
t2+12 +
12
∫dt√t2+1
= t√
t2+12 +
ln (t+√
t2+1)2 +c =
(x+2)√
(x+2)2+12 +
ln (x+2+√
(x+2)2+1)2 +c
= (x+2)√
x2+4x+52 +
ln (x+2+√
x2+4x+5)2 +c, x ∈R, c∈R.∫ √
x2+4x +3 dx = (x+2)√
x2+4x+32 −
ln |x+2+√
x2+4x+3|2 +c
=[ Subst. t =x +2 x2+4x +3=x2+4x +4−1 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx =(x +2)2−1= t2−1 x ∈(−1;∞) , t∈(1;∞)
]=∫ √
t2−1 dt
= t√
t2−12 −
12
∫dt√t2−1
= t√
t2−12 −
ln |t+√
t2−1|2 +c
= (x+2)√
x2+4x+32 −
ln |x+2+√
x2+4x+3|2 +c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 115, 116∫ √x2+4x +5 dx = (x+2)
√x2+4x+52 +
ln (x+2+√
x2+4x+5)2 +c
=[ Subst. t =x +2 x2+4x +5=x2+4x +4+1 x ∈R
dt =dx =(x +2)2+1= t2+1 t∈R
]=∫ √
t2+1 dt = t√
t2+12 +
12
∫dt√t2+1
= t√
t2+12 +
ln (t+√
t2+1)2 +c =
(x+2)√
(x+2)2+12 +
ln (x+2+√
(x+2)2+1)2 +c
= (x+2)√
x2+4x+52 +
ln (x+2+√
x2+4x+5)2 +c, x ∈R, c∈R.∫ √
x2+4x +3 dx = (x+2)√
x2+4x+32 −
ln |x+2+√
x2+4x+3|2 +c
=[ Subst. t =x +2 x2+4x +3=x2+4x +4−1 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx =(x +2)2−1= t2−1 x ∈(−1;∞) , t∈(1;∞)
]=∫ √
t2−1 dt
= t√
t2−12 −
12
∫dt√t2−1
= t√
t2−12 −
ln |t+√
t2−1|2 +c
= (x+2)√
x2+4x+32 −
ln |x+2+√
x2+4x+3|2 +c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 115, 116∫ √x2+4x +5 dx = (x+2)
√x2+4x+52 +
ln (x+2+√
x2+4x+5)2 +c
=[ Subst. t =x +2 x2+4x +5=x2+4x +4+1 x ∈R
dt =dx =(x +2)2+1= t2+1 t∈R
]=∫ √
t2+1 dt = t√
t2+12 +
12
∫dt√t2+1
= t√
t2+12 +
ln (t+√
t2+1)2 +c =
(x+2)√
(x+2)2+12 +
ln (x+2+√
(x+2)2+1)2 +c
= (x+2)√
x2+4x+52 +
ln (x+2+√
x2+4x+5)2 +c, x ∈R, c∈R.∫ √
x2+4x +3 dx = (x+2)√
x2+4x+32 −
ln |x+2+√
x2+4x+3|2 +c
=[ Subst. t =x +2 x2+4x +3=x2+4x +4−1 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx =(x +2)2−1= t2−1 x ∈(−1;∞) , t∈(1;∞)
]=∫ √
t2−1 dt
= t√
t2−12 −
12
∫dt√t2−1
= t√
t2−12 −
ln |t+√
t2−1|2 +c
= (x+2)√
x2+4x+32 −
ln |x+2+√
x2+4x+3|2 +c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 115, 116∫ √x2+4x +5 dx = (x+2)
√x2+4x+52 +
ln (x+2+√
x2+4x+5)2 +c
=[ Subst. t =x +2 x2+4x +5=x2+4x +4+1 x ∈R
dt =dx =(x +2)2+1= t2+1 t∈R
]=∫ √
t2+1 dt = t√
t2+12 +
12
∫dt√t2+1
= t√
t2+12 +
ln (t+√
t2+1)2 +c =
(x+2)√
(x+2)2+12 +
ln (x+2+√
(x+2)2+1)2 +c
= (x+2)√
x2+4x+52 +
ln (x+2+√
x2+4x+5)2 +c, x ∈R, c∈R.∫ √
x2+4x +3 dx = (x+2)√
x2+4x+32 −
ln |x+2+√
x2+4x+3|2 +c
=[ Subst. t =x +2 x2+4x +3=x2+4x +4−1 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx =(x +2)2−1= t2−1 x ∈(−1;∞) , t∈(1;∞)
]=∫ √
t2−1 dt
= t√
t2−12 −
12
∫dt√t2−1
= t√
t2−12 −
ln |t+√
t2−1|2 +c
= (x+2)√
x2+4x+32 −
ln |x+2+√
x2+4x+3|2 +c, x ∈(−∞;−3) ∪ (−1;∞), c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 117, 118∫ √x2−4x +6 dx = (x−2)
√x2−4x+62 +
ln (x−2+√
x2−4x+6)2 +c
=[Subst. t =x−2 x2−4x +6=x2+4x +4+2 x ∈R
dt =dx =(x−2)2+2= t2+(√2)2 t∈R
]=∫√
t2+(√2)2 dt
= t√
t2+(√2)2
2 +12
∫dt√
t2+(√2)2
= t√
t2+(√2)2
2 +ln (t+√
t2+(√2)2)
2 +c
= (x−2)√
x2−4x+62 +
ln (x−2+√
x2−4x+6)2 +c, x ∈R, c∈R.∫ √
x2−4x +2 dx = (x−2)√
x2+4x+22 −
ln |x−2+√
x2−4x+2|2 +c
=[Subst. t =x−2 x2−4x +2=x2−4x +4−2 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx =(x−2)2−2= t2−(√2)2 x ∈(−1;∞) , t∈(1;∞)
]=∫ √
t2−(√2)2 dt
= t√
t2−(√2)2
2 −12
∫dt√
t2−(√2)2
= t√
t2−(√2)2
2 −ln∣∣t+√t2−(√2)2∣∣
2 +c
= (x−2)√
x2+4x+22 −
ln |x−2+√
x2−4x+2|2 +c, x ∈R−
〈2−√2; 2+
√2〉, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 117, 118∫ √x2−4x +6 dx = (x−2)
√x2−4x+62 +
ln (x−2+√
x2−4x+6)2 +c
=[Subst. t =x−2 x2−4x +6=x2+4x +4+2 x ∈R
dt =dx =(x−2)2+2= t2+(√2)2 t∈R
]=∫√
t2+(√2)2 dt
= t√
t2+(√2)2
2 +12
∫dt√
t2+(√2)2
= t√
t2+(√2)2
2 +ln (t+√
t2+(√2)2)
2 +c
= (x−2)√
x2−4x+62 +
ln (x−2+√
x2−4x+6)2 +c, x ∈R, c∈R.∫ √
x2−4x +2 dx = (x−2)√
x2+4x+22 −
ln |x−2+√
x2−4x+2|2 +c
=[Subst. t =x−2 x2−4x +2=x2−4x +4−2 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx =(x−2)2−2= t2−(√2)2 x ∈(−1;∞) , t∈(1;∞)
]=∫ √
t2−(√2)2 dt
= t√
t2−(√2)2
2 −12
∫dt√
t2−(√2)2
= t√
t2−(√2)2
2 −ln∣∣t+√t2−(√2)2∣∣
2 +c
= (x−2)√
x2+4x+22 −
ln |x−2+√
x2−4x+2|2 +c, x ∈R−
〈2−√2; 2+
√2〉, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 117, 118∫ √x2−4x +6 dx = (x−2)
√x2−4x+62 +
ln (x−2+√
x2−4x+6)2 +c
=[Subst. t =x−2 x2−4x +6=x2+4x +4+2 x ∈R
dt =dx =(x−2)2+2= t2+(√2)2 t∈R
]=∫√
t2+(√2)2 dt
= t√
t2+(√2)2
2 +12
∫dt√
t2+(√2)2
= t√
t2+(√2)2
2 +ln (t+√
t2+(√2)2)
2 +c
= (x−2)√
x2−4x+62 +
ln (x−2+√
x2−4x+6)2 +c, x ∈R, c∈R.∫ √
x2−4x +2 dx = (x−2)√
x2+4x+22 −
ln |x−2+√
x2−4x+2|2 +c
=[Subst. t =x−2 x2−4x +2=x2−4x +4−2 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx =(x−2)2−2= t2−(√2)2 x ∈(−1;∞) , t∈(1;∞)
]=∫ √
t2−(√2)2 dt
= t√
t2−(√2)2
2 −12
∫dt√
t2−(√2)2
= t√
t2−(√2)2
2 −ln∣∣t+√t2−(√2)2∣∣
2 +c
= (x−2)√
x2+4x+22 −
ln |x−2+√
x2−4x+2|2 +c, x ∈R−
〈2−√2; 2+
√2〉, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 117, 118∫ √x2−4x +6 dx = (x−2)
√x2−4x+62 +
ln (x−2+√
x2−4x+6)2 +c
=[Subst. t =x−2 x2−4x +6=x2+4x +4+2 x ∈R
dt =dx =(x−2)2+2= t2+(√2)2 t∈R
]=∫√
t2+(√2)2 dt
= t√
t2+(√2)2
2 +12
∫dt√
t2+(√2)2
= t√
t2+(√2)2
2 +ln (t+√
t2+(√2)2)
2 +c
= (x−2)√
x2−4x+62 +
ln (x−2+√
x2−4x+6)2 +c, x ∈R, c∈R.∫ √
x2−4x +2 dx = (x−2)√
x2+4x+22 −
ln |x−2+√
x2−4x+2|2 +c
=[Subst. t =x−2 x2−4x +2=x2−4x +4−2 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx =(x−2)2−2= t2−(√2)2 x ∈(−1;∞) , t∈(1;∞)
]=∫ √
t2−(√2)2 dt
= t√
t2−(√2)2
2 −12
∫dt√
t2−(√2)2
= t√
t2−(√2)2
2 −ln∣∣t+√t2−(√2)2∣∣
2 +c
= (x−2)√
x2+4x+22 −
ln |x−2+√
x2−4x+2|2 +c, x ∈R−
〈2−√2; 2+
√2〉, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 117, 118∫ √x2−4x +6 dx = (x−2)
√x2−4x+62 +
ln (x−2+√
x2−4x+6)2 +c
=[Subst. t =x−2 x2−4x +6=x2+4x +4+2 x ∈R
dt =dx =(x−2)2+2= t2+(√2)2 t∈R
]=∫√
t2+(√2)2 dt
= t√
t2+(√2)2
2 +12
∫dt√
t2+(√2)2
= t√
t2+(√2)2
2 +ln (t+√
t2+(√2)2)
2 +c
= (x−2)√
x2−4x+62 +
ln (x−2+√
x2−4x+6)2 +c, x ∈R, c∈R.∫ √
x2−4x +2 dx = (x−2)√
x2+4x+22 −
ln |x−2+√
x2−4x+2|2 +c
=[Subst. t =x−2 x2−4x +2=x2−4x +4−2 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx =(x−2)2−2= t2−(√2)2 x ∈(−1;∞) , t∈(1;∞)
]=∫ √
t2−(√2)2 dt
= t√
t2−(√2)2
2 −12
∫dt√
t2−(√2)2
= t√
t2−(√2)2
2 −ln∣∣t+√t2−(√2)2∣∣
2 +c
= (x−2)√
x2+4x+22 −
ln |x−2+√
x2−4x+2|2 +c, x ∈R−
〈2−√2; 2+
√2〉, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 117, 118∫ √x2−4x +6 dx = (x−2)
√x2−4x+62 +
ln (x−2+√
x2−4x+6)2 +c
=[Subst. t =x−2 x2−4x +6=x2+4x +4+2 x ∈R
dt =dx =(x−2)2+2= t2+(√2)2 t∈R
]=∫√
t2+(√2)2 dt
= t√
t2+(√2)2
2 +12
∫dt√
t2+(√2)2
= t√
t2+(√2)2
2 +ln (t+√
t2+(√2)2)
2 +c
= (x−2)√
x2−4x+62 +
ln (x−2+√
x2−4x+6)2 +c, x ∈R, c∈R.∫ √
x2−4x +2 dx = (x−2)√
x2+4x+22 −
ln |x−2+√
x2−4x+2|2 +c
=[Subst. t =x−2 x2−4x +2=x2−4x +4−2 x ∈(−∞;−3) , t∈(−∞;−1)
dt =dx =(x−2)2−2= t2−(√2)2 x ∈(−1;∞) , t∈(1;∞)
]=∫ √
t2−(√2)2 dt
= t√
t2−(√2)2
2 −12
∫dt√
t2−(√2)2
= t√
t2−(√2)2
2 −ln∣∣t+√t2−(√2)2∣∣
2 +c
= (x−2)√
x2+4x+22 −
ln |x−2+√
x2−4x+2|2 +c, x ∈R−
〈2−√2; 2+
√2〉, c∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 119, 120∫ √−x2+4x−3 dx = (x−2)
√−x2+4x−32 +
arcsin (x−2)2 +c1
=[ Subst. t =x−2 −x2+4x−3=−(x2−4x +4−1) x ∈(1; 3)
dt =dx =−(x−2)2+1=1−t2 t∈(−1; 1)
]=∫ √
1−t2 dt
= t√
1−t22 +
12
∫dt√1−t2
= t√
1−t22 +
arcsin t2 +c1=
t√
1−t22 −
arccos t2 +c2
= (x−2)√−x2+4x−32 +
arcsin (x−2)2 +c1=
(x−2)√−x2+4x−32 −
arccos (x−2)2 +c2,
x ∈(1; 3), c1, c2∈R.∫ √x(2−x) dx =
∫ √−x2+2x dx = (x−2)
√−x2+2x2 +
arcsin (x−2)2 +c1
=[ Subst. t =x−1 −x2+2x =−(x2−2x +1−1) x ∈(0; 2)
dt =dx =−(x−1)2+1=1−t2 t∈(−1; 1)
]=∫ √
1−t2 dt = t√
1−t22 +
12
∫dt√1−t2
= t√
1−t22 +
arcsin t2 +c1=
t√
1−t22 −
arccos t2 +c2=
(x−2)√−x2+2x2 +
arcsin (x−2)2 +c1
= (x−2)√−x2+2x2 −
arccos (x−2)2 +c2, x ∈(0; 2), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 119, 120∫ √−x2+4x−3 dx = (x−2)
√−x2+4x−32 +
arcsin (x−2)2 +c1
=[ Subst. t =x−2 −x2+4x−3=−(x2−4x +4−1) x ∈(1; 3)
dt =dx =−(x−2)2+1=1−t2 t∈(−1; 1)
]=∫ √
1−t2 dt
= t√
1−t22 +
12
∫dt√1−t2
= t√
1−t22 +
arcsin t2 +c1=
t√
1−t22 −
arccos t2 +c2
= (x−2)√−x2+4x−32 +
arcsin (x−2)2 +c1=
(x−2)√−x2+4x−32 −
arccos (x−2)2 +c2,
x ∈(1; 3), c1, c2∈R.∫ √x(2−x) dx =
∫ √−x2+2x dx = (x−2)
√−x2+2x2 +
arcsin (x−2)2 +c1
=[ Subst. t =x−1 −x2+2x =−(x2−2x +1−1) x ∈(0; 2)
dt =dx =−(x−1)2+1=1−t2 t∈(−1; 1)
]=∫ √
1−t2 dt = t√
1−t22 +
12
∫dt√1−t2
= t√
1−t22 +
arcsin t2 +c1=
t√
1−t22 −
arccos t2 +c2=
(x−2)√−x2+2x2 +
arcsin (x−2)2 +c1
= (x−2)√−x2+2x2 −
arccos (x−2)2 +c2, x ∈(0; 2), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 119, 120∫ √−x2+4x−3 dx = (x−2)
√−x2+4x−32 +
arcsin (x−2)2 +c1
=[ Subst. t =x−2 −x2+4x−3=−(x2−4x +4−1) x ∈(1; 3)
dt =dx =−(x−2)2+1=1−t2 t∈(−1; 1)
]=∫ √
1−t2 dt
= t√
1−t22 +
12
∫dt√1−t2
= t√
1−t22 +
arcsin t2 +c1=
t√
1−t22 −
arccos t2 +c2
= (x−2)√−x2+4x−32 +
arcsin (x−2)2 +c1=
(x−2)√−x2+4x−32 −
arccos (x−2)2 +c2,
x ∈(1; 3), c1, c2∈R.∫ √x(2−x) dx =
∫ √−x2+2x dx = (x−2)
√−x2+2x2 +
arcsin (x−2)2 +c1
=[ Subst. t =x−1 −x2+2x =−(x2−2x +1−1) x ∈(0; 2)
dt =dx =−(x−1)2+1=1−t2 t∈(−1; 1)
]=∫ √
1−t2 dt = t√
1−t22 +
12
∫dt√1−t2
= t√
1−t22 +
arcsin t2 +c1=
t√
1−t22 −
arccos t2 +c2=
(x−2)√−x2+2x2 +
arcsin (x−2)2 +c1
= (x−2)√−x2+2x2 −
arccos (x−2)2 +c2, x ∈(0; 2), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 119, 120∫ √−x2+4x−3 dx = (x−2)
√−x2+4x−32 +
arcsin (x−2)2 +c1
=[ Subst. t =x−2 −x2+4x−3=−(x2−4x +4−1) x ∈(1; 3)
dt =dx =−(x−2)2+1=1−t2 t∈(−1; 1)
]=∫ √
1−t2 dt
= t√
1−t22 +
12
∫dt√1−t2
= t√
1−t22 +
arcsin t2 +c1=
t√
1−t22 −
arccos t2 +c2
= (x−2)√−x2+4x−32 +
arcsin (x−2)2 +c1=
(x−2)√−x2+4x−32 −
arccos (x−2)2 +c2,
x ∈(1; 3), c1, c2∈R.∫ √x(2−x) dx =
∫ √−x2+2x dx = (x−2)
√−x2+2x2 +
arcsin (x−2)2 +c1
=[ Subst. t =x−1 −x2+2x =−(x2−2x +1−1) x ∈(0; 2)
dt =dx =−(x−1)2+1=1−t2 t∈(−1; 1)
]=∫ √
1−t2 dt = t√
1−t22 +
12
∫dt√1−t2
= t√
1−t22 +
arcsin t2 +c1=
t√
1−t22 −
arccos t2 +c2=
(x−2)√−x2+2x2 +
arcsin (x−2)2 +c1
= (x−2)√−x2+2x2 −
arccos (x−2)2 +c2, x ∈(0; 2), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 119, 120∫ √−x2+4x−3 dx = (x−2)
√−x2+4x−32 +
arcsin (x−2)2 +c1
=[ Subst. t =x−2 −x2+4x−3=−(x2−4x +4−1) x ∈(1; 3)
dt =dx =−(x−2)2+1=1−t2 t∈(−1; 1)
]=∫ √
1−t2 dt
= t√
1−t22 +
12
∫dt√1−t2
= t√
1−t22 +
arcsin t2 +c1=
t√
1−t22 −
arccos t2 +c2
= (x−2)√−x2+4x−32 +
arcsin (x−2)2 +c1=
(x−2)√−x2+4x−32 −
arccos (x−2)2 +c2,
x ∈(1; 3), c1, c2∈R.∫ √x(2−x) dx =
∫ √−x2+2x dx = (x−2)
√−x2+2x2 +
arcsin (x−2)2 +c1
=[ Subst. t =x−1 −x2+2x =−(x2−2x +1−1) x ∈(0; 2)
dt =dx =−(x−1)2+1=1−t2 t∈(−1; 1)
]=∫ √
1−t2 dt = t√
1−t22 +
12
∫dt√1−t2
= t√
1−t22 +
arcsin t2 +c1=
t√
1−t22 −
arccos t2 +c2=
(x−2)√−x2+2x2 +
arcsin (x−2)2 +c1
= (x−2)√−x2+2x2 −
arccos (x−2)2 +c2, x ∈(0; 2), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 11–12 13–14 15–16 17–18 19–20
Riešené príklady – 119, 120∫ √−x2+4x−3 dx = (x−2)
√−x2+4x−32 +
arcsin (x−2)2 +c1
=[ Subst. t =x−2 −x2+4x−3=−(x2−4x +4−1) x ∈(1; 3)
dt =dx =−(x−2)2+1=1−t2 t∈(−1; 1)
]=∫ √
1−t2 dt
= t√
1−t22 +
12
∫dt√1−t2
= t√
1−t22 +
arcsin t2 +c1=
t√
1−t22 −
arccos t2 +c2
= (x−2)√−x2+4x−32 +
arcsin (x−2)2 +c1=
(x−2)√−x2+4x−32 −
arccos (x−2)2 +c2,
x ∈(1; 3), c1, c2∈R.∫ √x(2−x) dx =
∫ √−x2+2x dx = (x−2)
√−x2+2x2 +
arcsin (x−2)2 +c1
=[ Subst. t =x−1 −x2+2x =−(x2−2x +1−1) x ∈(0; 2)
dt =dx =−(x−1)2+1=1−t2 t∈(−1; 1)
]=∫ √
1−t2 dt = t√
1−t22 +
12
∫dt√1−t2
= t√
1−t22 +
arcsin t2 +c1=
t√
1−t22 −
arccos t2 +c2=
(x−2)√−x2+2x2 +
arcsin (x−2)2 +c1
= (x−2)√−x2+2x2 −
arccos (x−2)2 +c2, x ∈(0; 2), c1, c2∈R.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 21 22 23–24 25 26 27–28 29 30
Riešené príklady – 121
In =∫
dx(x2+a2)n =
2n−32a2(n−1) In−1+
x2a2(n−1)(x2+a2)n−1 , I1=
1a arctg
xa +c, a>0, n∈N
= 1a2∫
a2 dx(x2+a2)n =
1a2
∫(x2+a2−x2) dx
(x2+a2)n =1a2
∫(x2+a2) dx(x2+a2)n −
1a2
∫x2 dx
(x2+a2)n
= 1a2∫
dx(x2+a2)n−1 −
12a2
∫x ·2x dx
(x2+a2)n =[
u = x u′= 1v ′= 2x(x2+a2)n v =
(x2+a2)1−n1−n =
1(1−n)(x2+a2)n−1
]
= 1a2 In−1 −12a2
[x
(1−n)(x2+a2)n−1 −1
1−n
∫dx
(x2+a2)n−1
]= 1a2 In−1 −
x2a2(1−n)(x2+a2)n−1 +
12a2(1−n) In−1
= 12a2 (2+1
1−n )In−1 +x
2a2(n−1)(x2+a2)n−1 =3−2n
2a2(1−n) In−1 +x
2a2(n−1)(x2+a2)n−1
= 2n−32a2(n−1) In−1 +x
2a2(n−1)(x2+a2)n−1 , x ∈R, c∈R, n=2,3,4, . . . .
I1=∫
dxx2+a2 =
1a arctg
xa + c, x ∈R, c∈R, n=1.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 21 22 23–24 25 26 27–28 29 30
Riešené príklady – 121
In =∫
dx(x2+a2)n =
2n−32a2(n−1) In−1+
x2a2(n−1)(x2+a2)n−1 , I1=
1a arctg
xa +c, a>0, n∈N
= 1a2∫
a2 dx(x2+a2)n =
1a2
∫(x2+a2−x2) dx
(x2+a2)n =1a2
∫(x2+a2) dx(x2+a2)n −
1a2
∫x2 dx
(x2+a2)n
= 1a2∫
dx(x2+a2)n−1 −
12a2
∫x ·2x dx
(x2+a2)n =[
u = x u′= 1v ′= 2x(x2+a2)n v =
(x2+a2)1−n1−n =
1(1−n)(x2+a2)n−1
]
= 1a2 In−1 −12a2
[x
(1−n)(x2+a2)n−1 −1
1−n
∫dx
(x2+a2)n−1
]= 1a2 In−1 −
x2a2(1−n)(x2+a2)n−1 +
12a2(1−n) In−1
= 12a2 (2+1
1−n )In−1 +x
2a2(n−1)(x2+a2)n−1 =3−2n
2a2(1−n) In−1 +x
2a2(n−1)(x2+a2)n−1
= 2n−32a2(n−1) In−1 +x
2a2(n−1)(x2+a2)n−1 , x ∈R, c∈R, n=2,3,4, . . . .
I1=∫
dxx2+a2 =
1a arctg
xa + c, x ∈R, c∈R, n=1.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 21 22 23–24 25 26 27–28 29 30
Riešené príklady – 121
In =∫
dx(x2+a2)n =
2n−32a2(n−1) In−1+
x2a2(n−1)(x2+a2)n−1 , I1=
1a arctg
xa +c, a>0, n∈N
= 1a2∫
a2 dx(x2+a2)n =
1a2
∫(x2+a2−x2) dx
(x2+a2)n =1a2
∫(x2+a2) dx(x2+a2)n −
1a2
∫x2 dx
(x2+a2)n
= 1a2∫
dx(x2+a2)n−1 −
12a2
∫x ·2x dx
(x2+a2)n =[
u = x u′= 1v ′= 2x(x2+a2)n v =
(x2+a2)1−n1−n =
1(1−n)(x2+a2)n−1
]
= 1a2 In−1 −12a2
[x
(1−n)(x2+a2)n−1 −1
1−n
∫dx
(x2+a2)n−1
]= 1a2 In−1 −
x2a2(1−n)(x2+a2)n−1 +
12a2(1−n) In−1
= 12a2 (2+1
1−n )In−1 +x
2a2(n−1)(x2+a2)n−1 =3−2n
2a2(1−n) In−1 +x
2a2(n−1)(x2+a2)n−1
= 2n−32a2(n−1) In−1 +x
2a2(n−1)(x2+a2)n−1 , x ∈R, c∈R, n=2,3,4, . . . .
I1=∫
dxx2+a2 =
1a arctg
xa + c, x ∈R, c∈R, n=1.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 21 22 23–24 25 26 27–28 29 30
Riešené príklady – 121
In =∫
dx(x2+a2)n =
2n−32a2(n−1) In−1+
x2a2(n−1)(x2+a2)n−1 , I1=
1a arctg
xa +c, a>0, n∈N
= 1a2∫
a2 dx(x2+a2)n =
1a2
∫(x2+a2−x2) dx
(x2+a2)n =1a2
∫(x2+a2) dx(x2+a2)n −
1a2
∫x2 dx
(x2+a2)n
= 1a2∫
dx(x2+a2)n−1 −
12a2
∫x ·2x dx
(x2+a2)n =[
u = x u′= 1v ′= 2x(x2+a2)n v =
(x2+a2)1−n1−n =
1(1−n)(x2+a2)n−1
]
= 1a2 In−1 −12a2
[x
(1−n)(x2+a2)n−1 −1
1−n
∫dx
(x2+a2)n−1
]= 1a2 In−1 −
x2a2(1−n)(x2+a2)n−1 +
12a2(1−n) In−1
= 12a2 (2+1
1−n )In−1 +x
2a2(n−1)(x2+a2)n−1 =3−2n
2a2(1−n) In−1 +x
2a2(n−1)(x2+a2)n−1
= 2n−32a2(n−1) In−1 +x
2a2(n−1)(x2+a2)n−1 , x ∈R, c∈R, n=2,3,4, . . . .
I1=∫
dxx2+a2 =
1a arctg
xa + c, x ∈R, c∈R, n=1.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 21 22 23–24 25 26 27–28 29 30
Riešené príklady – 121
In =∫
dx(x2+a2)n =
2n−32a2(n−1) In−1+
x2a2(n−1)(x2+a2)n−1 , I1=
1a arctg
xa +c, a>0, n∈N
= 1a2∫
a2 dx(x2+a2)n =
1a2
∫(x2+a2−x2) dx
(x2+a2)n =1a2
∫(x2+a2) dx(x2+a2)n −
1a2
∫x2 dx
(x2+a2)n
= 1a2∫
dx(x2+a2)n−1 −
12a2
∫x ·2x dx
(x2+a2)n =[
u = x u′= 1v ′= 2x(x2+a2)n v =
(x2+a2)1−n1−n =
1(1−n)(x2+a2)n−1
]
= 1a2 In−1 −12a2
[x
(1−n)(x2+a2)n−1 −1
1−n
∫dx
(x2+a2)n−1
]= 1a2 In−1 −
x2a2(1−n)(x2+a2)n−1 +
12a2(1−n) In−1
= 12a2 (2+1
1−n )In−1 +x
2a2(n−1)(x2+a2)n−1 =3−2n
2a2(1−n) In−1 +x
2a2(n−1)(x2+a2)n−1
= 2n−32a2(n−1) In−1 +x
2a2(n−1)(x2+a2)n−1 , x ∈R, c∈R, n=2,3,4, . . . .
I1=∫
dxx2+a2 =
1a arctg
xa + c, x ∈R, c∈R, n=1.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 21 22 23–24 25 26 27–28 29 30
Riešené príklady – 121
In =∫
dx(x2+a2)n =
2n−32a2(n−1) In−1+
x2a2(n−1)(x2+a2)n−1 , I1=
1a arctg
xa +c, a>0, n∈N
= 1a2∫
a2 dx(x2+a2)n =
1a2
∫(x2+a2−x2) dx
(x2+a2)n =1a2
∫(x2+a2) dx(x2+a2)n −
1a2
∫x2 dx
(x2+a2)n
= 1a2∫
dx(x2+a2)n−1 −
12a2
∫x ·2x dx
(x2+a2)n =[
u = x u′= 1v ′= 2x(x2+a2)n v =
(x2+a2)1−n1−n =
1(1−n)(x2+a2)n−1
]
= 1a2 In−1 −12a2
[x
(1−n)(x2+a2)n−1 −1
1−n
∫dx
(x2+a2)n−1
]= 1a2 In−1 −
x2a2(1−n)(x2+a2)n−1 +
12a2(1−n) In−1
= 12a2 (2+1
1−n )In−1 +x
2a2(n−1)(x2+a2)n−1 =3−2n
2a2(1−n) In−1 +x
2a2(n−1)(x2+a2)n−1
= 2n−32a2(n−1) In−1 +x
2a2(n−1)(x2+a2)n−1 , x ∈R, c∈R, n=2,3,4, . . . .
I1=∫
dxx2+a2 =
1a arctg
xa + c, x ∈R, c∈R, n=1.
[email protected] http://frcatel.fri.uniza.sk/˜beerb
mailto:[email protected]://frcatel.fri.uniza.sk/~beerb
-
Obsah Zoznam 01 11 21 31 41 51 61 71 81 91 21 22 23–24 25 26 27–28 29 30
Riešené príklady – 121
In =∫
dx(x2+a2)n =
2n−32a2(n−1) In−1+
x2a2(n−1)(x2+a2)n−1 , I1=
1a arctg
xa +c, a>0, n∈N
= 1a2∫
a2 dx(x2+a2)n =
1a2
∫(x2+a2−x2) dx
(x2+a2)n =1a2
∫(x2+a2) dx(x2+a2)n −
1a2
∫x2 dx
(x2+a2)n
= 1a2∫
dx(x2+a2)n−1 −
12a2
∫x ·2x dx
(x2+a2)n =[
u = x u′= 1v ′= 2x(x2+a2)n v =
(x2+a2)1−n1−n =
1(1−n)(x2+a2)n−1
]
= 1a2 In−1 −12a2
[x
(1−n)(x2+a2)n−1 −1
1−n
∫dx
(x2+a2)n−1
]= 1a2 In−1 −
x2a2(1−n)(x2+a2)n−1 +
12a2(1−n) In−1
= 12a2 (2+1
1−n )In−1 +x
2a2(n−1)(x2+a2)n−1 =3−2n
2a2(1−n)