Integration e.g.

8/3/2019 Integration e.g.

http://slidepdf.com/reader/full/integration-eg 1/11

Examples In Integration

2 2I x a x dx= −∫ let

2 2 du duu a x 2x dx

dx 2x= − ⇒ = − ⇒ =

−, then

duI x u

2x= ⋅

−∫

3

1 3 322 22 2 2

1 1 u 1 1u du c u c (a x ) c

32 2 3 32

= − = − + = − + = − − +

∫

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2

2 3

xI dx

a x=

+∫ let

2 3 2

2

du duu a x 3x dx

dx 3x= + ⇒ = ⇒ = , then

2

2

x duI

3xu= ⋅∫

11 1 12

2 32 2 21 1 u 2 2

u du c u c (a x ) c13 3 3 32

− = = + = + = + +

∫

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2I sin xcosxdx= ∫ let2 du

u sin x 2cosxsinxdx

= ⇒ = &dv

cosx v sinxdx

= ⇒ = , then

33 2 3 3 sin x

I sin x 2 sin xcosxdx sin x 2I 3I sin x I c3

= − = − ⇒ = ⇒ = +∫ -----------------------------------------------------------------------------------

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3xI dxx 1= +∫ , by algebraic long division

3

2

x 1x x 1x 1 x 1= − + −+ + , then this

implies that2 3

2 1 x xI 1dx xdx xdx dx x ln(x 1) c

x 1 2 3= − + − = − + − + +

+∫ ∫ ∫ ∫ -----------------------------------------------------------------------------------

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2 2sec 1 3sec 1I d d ln(1 3tan ) c

1 3tan 3 1 3tan 3

θ θ= θ = θ = + θ +

+ θ + θ∫ ∫ , sincef'(x)

dx ln(f(x)) cf(x)

= +∫ -----------------------------------------------------------------------------------

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2

secx(secx tanx) sec x secxtanxI secxdx dx dx ln(secx tanx) csecx tanx secx tanx

+ += = = = + ++ +∫ ∫ ∫

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1I dx

1 cosx=

+∫ let 2

x 2t tan dx dt

2 1 t

= ⇒ = + , then

2

2

2

2

1 tI dt1 t

11 t

+=−

++

∫

2

2 2

2 2 x 1 xdt dt 1dt t c tan c tanx 1 tan c

1 t 1 t 2 2 2 2

= = = = + = + = − + + + − ∫ ∫ ∫ , from



2

2tanAtan2A

1 tan A=

−,

2

2

xsin

1 1 1 cosx2I tan 1 c tanx 1 c

x2 2 1 cosxcos

2

− = − + = − + +

, from

2 2

cos2A 2cos A 1 1 2sin A= − = − , so now we have1 1 cosx 1 cosx

I tanx c2 1 cosx 1 cosx

+ − = − + + + 1 2cosx cosx sinx cosx sinxtanx c tanx c c c

2 1 cosx 1 cosx cosx 1 cosx 1 cosx

= + = + = + = + + + + +

2 2

sinx(1 cosx) sinx sinxcosx sinx(1 cosx) 1 cosxc c c c

(1 cosx)(1 cosx) 1 cos x sin x sinx

− − − −= + = + = + = +

+ − −,

therefore we finally have,1 cosx

I c cotx cosecx csinx sinx

= − + = − + +

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2 2

1I dx

x a=

+∫ let 2x atan dx asec d= θ ⇒ = θ θ , then2

2 2 2

asecI d

a tan a

θ= θ

θ +∫

2 21

2 2 2

asec 1 sec 1 1 1 xd d 1d c tan c

a (tan 1) a sec a a a a

−θ θ = θ = θ = θ = θ + = + θ + θ ∫ ∫ ∫ -----------------------------------------------------------------------------------

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2 2

1I dx

a x=

−

∫ let x asin dx acos d= θ ⇒ = θ θ , then2 2 2

acosI d

a a sin

θ= θ

− θ

∫

1

2 2

acos cos cos xd d d 1d c sin c

cos aa 1 sin cos

−θ θ θ = θ = θ = θ = θ = θ + = + θ − θ θ∫ ∫ ∫ ∫

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2 2

1I dx

x a=

−∫ let x asec dx asec tan d= θ ⇒ = θ θ θ , then 2 2 2

asec tanI d

a sec a

θ θ= θ

θ −∫

2 2 2

asec tan 1 sec tan 1 sec 1 1d d d d

a (sec 1) a tan a tan a sin

θ θ θ θ θ= θ = θ = θ = θ

θ − θ θ θ∫ ∫ ∫ ∫ let t tan2

θ =

22d dt

1 t⇒ θ =

+, so 2

2

1 1 2 1 1 1 1I dt dt lnt c ln tan c2ta 1 t a t a a 2

1 t

θ = ⋅ = = + = + +

+∫ ∫

2

2

2

sin1 1 1 1 2ln tan 1 tan c ln tan 1 c

a 2 2 a 2cos

2

θ θ = θ − + = θ − + θ

1 1 1 cos 1 1 1 cos 1 cosln tan 1 c tan c

a 2 1 cos a 2 1 cos 1 cos

− θ + θ − θ = θ − + = θ − + + θ + θ + θ

1 1 2cos 1 cos 1 sin cosln tan c ln tan c ln ca 2 1 cos a 1 cos a cos 1 cos

θ θ θ θ = θ + = θ + = ⋅ + + θ + θ θ + θ



21 sin 1 tan cos 1 sec 1 cosln c ln c ln c

a 1 cos a 1 cos a 1 cos

θ θ θ θ − ⋅ θ = + = + = + + θ + θ + θ

( ) ( ){ }2

21 sec 1 1ln c ln sec 1 ln sec 1 c

a sec 1 a

θ −= + = θ − − θ + +

θ +

( ) ( ) ( ) ( )( ) ( )21 1 1 1ln sec 1 ln sec 1 c ln sec 1 sec 1 ln sec 1 c

a 2 a 2

= θ − − θ + + = θ − θ + − θ + +

( ) ( )( ) ( ){ }1 x x x 1

ln 1 1 2ln 1 c ln x a x a 2ln x a c2a a a a 2a

= − + − + + = − + − + +

( ) ( )

( )2

x a x a1ln c

2a x a

− + = + +

, we finally have 1 x a

I ln c2a x a

− = + +

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2 2

1I dx

x a=

+∫ let 2x atan dx asec d= θ ⇒ = θ θ , then

2

2 2 2

asecI d

a tan a

θ= θ

θ +∫

( )2 2

2

asec secd d sec d ln tan sec c

seca tan 1

θ θ= θ = θ = θ θ = θ + θ +

θθ +∫ ∫ ∫

( ) ( )2

2 2 2

2

x xln tan tan 1 c ln 1 c ln x x a c

a a

= θ + θ + + = + + + = + + +

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2 2

1I dx

x a=

−∫ let x asec dx asec tan d= θ ⇒ = θ θ θ , then

2 2 2

asec tanI d

a sec a

θ θ= θ

θ −∫

( )2 2

asec tan sec tand d sec d ln sec tan c

a sec 1 tan

θ θ θ θ= θ = θ = θ θ = θ + θ +

θ − θ∫ ∫ ∫

( ) ( )2

2 2 2

2

x xln sec sec 1 c ln 1 c ln x x a c

a a

= θ + θ − + = + − + = + − +

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2

1I dx

x 2x 5=

+ +∫ completing the square,

1

2

1 1 x 1I dx tan c

(x 1) 4 2 2

− + = = + + + ∫ -----------------------------------------------------------------------------------

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2

1I dx

2 x x=

+ −∫ manipulate in order to complete the square,

2

1I dx

x x 2=

− − − ∫

1 1

22

1x

1 1 2x 12dx dx sin c sin c3 39 11 9

xx 24 22 4

− −

− − = = = + = + − −− − −

∫ ∫

2

sinI d

9 4cos

θ= θ

− θ∫ let

12cos x d

2sinθ = ⇒ θ = −

θ, then

2

1 sin dxI

2 sin9 x

θ= − ⋅

θ−∫

1 1

2

1 dx 1 x 1 2cossin c sin c

2 2 3 2 39 x

− − θ = − = − + = − + −

∫ -----------------------------------------------------------------------------------

-------------------------------------------

( )2

2 5 2 2 2 2 2I sin xcos xdx sin xcos xcos xcosxdx sin x 1 sin x cosxdx= = = −∫ ∫ ∫

( ) ( )2 2 4 2 4 6sin x 1 2sin x sin x cosxdx sin x 2sin x sin x cosxdx= − + = − +∫ ∫ let u sinx=

du ducosx dx

dx cosx⇒ = ⇒ = , then ( ) ( )2 4 6 2 4 6du

I u 2u u cosx u 2u u ducosx

= − + ⋅ = − +∫ ∫ ,

so by normal power integration,3 5 7 3 5 7u 2u u sin x 2sin x sin x

I c c

3 5 7 3 5 7

= − + + = − + +

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( )3 2 2I cos xdx cos xcosxdx 1 sin x cosxdx= = = −∫ ∫ ∫ , letdu

u sinx dxcosx

= ⇒ = , then

( ) ( )3 3

2 2du u sin xI 1 u cosx 1 u du u c sinx c

cosx 3 3= − ⋅ = − = − + = − +∫ ∫

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( ) ( )2 2 2 2 2 4I sin xcos xdx sin x 1 sin x dx sin x sin x dx= = − = −∫ ∫ ∫ , to integrate 4sin x we

need a formula with this included, so consider( )cos4x cos 2x 2x cos2xcos2x sin2xsin2x= + = −

( ) ( ) ( ) ( )2 2 2 4 2 21 2sin x 1 2sin x 2sinxcosx 2sinxcosx 1 4sin x 4sin x 4sin xcos x= − − − = − + −

( )2 4 2 2 2 4 2 41 4sin x 4sin x 4sin x 1 sin x 1 4sin x 4sin x 4sin x 4sin x= − + − − = − + − +

2 4 4 2 4 21 11 8sin x 8sin x 8sin x cos4x 1 8sin x sin x cos4x sin x

8 8= − + ⇒ = − + ⇒ = − + ,

( )2 21 1 1 x sin4xI sin x cos4x sin x dx 1 cos4x dx c

8 8 8 8 32

= − − + = − = − +

∫ ∫ -----------------------------------------------------------------------------------

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2I cos xdx= ∫ , ( 2cos2x 2cos x 1= − ) therefore, ( )

1 x 1I cos2x 1 dx sin2x c

2 2 4= + = + +∫

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2 2

I a x dx= −∫

letx asin dx acos d= θ ⇒ = θ θ

, then

2 2 2

I a a sin acos d= − θ ⋅ θ θ∫

( )2 2

2 2 2 2 a sin2 aa 1 sin cos d a cos d c sin cos c

2 2 2

θ = − θ θ θ = θ θ = + θ + = θ θ + θ + ∫ ∫

2 2 22 1 1

2

a x x a x x x1 sin sin c 1 sin c

2 a a 2 a a a

− − = ⋅ − θ + + = ⋅ − + +

2 22 2 1 2 2 1a x 1 x x a x

a x sin c a x sin c2 a a a 2 2 a

− − = ⋅ ⋅ − + + = − + +

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( )3

2 2 2

1I dx

a x

=−

∫ let x asin dx acos d= θ ⇒ = θ θ , then( )

32 2 2 2

acosI d

a a sin

θ= θ− θ

( ) ( )

2

3 32 2 2 2 2 22 22 2

1 cos 1 cos 1 1 1 tand d d sec c

a a a cos a a1 sin cos

θ θ θ= θ = θ = θ = θ = +

θ− θ θ

∫ ∫ ∫ ∫

2 3 2 2 2 2 23

2

sin x x xc c c c

a cos a 1 sin x a a xa 1

a

θ= + = + = + = +

θ − θ −−

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2

1I dx

x x 1=

+∫ let 2x tan dx sec d= θ ⇒ = θ θ , then

2

2

secI d

tan tan 1

θ= θ

θ θ +∫

2sec sec 1d d d

tan sec tan sin

θ θ= θ = θ = θ

θ θ θ θ∫ ∫ ∫ let 2

2t tan d dt

2 1 t

θ = ⇒ θ = + , then

2

2

1 2 2 1I dt dt dt lnt c ln tan c

2t 1 t 2t t 2

1 t

θ = ⋅ = = = + = + + +

∫ ∫ ∫ , now from double

angle formulae

2

2

2

sin1 1 2

I ln tan 1 tan c ln tan 1 c2 2 2

cos2

θ θ = θ − + = θ − + θ

1 1 cos 1 2cos cosln tan 1 c ln tan c ln tan c

2 1 cos 2 1 cos 1 cos

− θ θ θ = θ − + = θ + = θ + + θ + θ + θ



( )

( )

2

22 2 2

x x 1 1tan tan xln c ln c ln c ln c

sec 1 tan 1 1 x 1 1 x 1 1

+ − θ θ = + = + = + = + θ + θ + + + + + −

( )2

2

2

x x 1 1 x 1 1ln c ln cx x

+ − + − = + = + -----------------------------------------------------------------------------------

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3 2

2x 3I dx

x x 2x

+=

+ −∫ now if 3 2f(x) x x 2x= + − , then f(1) 0 (x 1)= ⇒ − is a

factor, so by long division 3 2 2x x 2x (x )(x 2x) (x 1)x(x 2)+ − = − + = − + , by

partial fractions we have that5 1 3 1 1 1 3 5 1

I dx dx dx lnx ln(x 1) ln(x 2) c3 x 1 2 x 6 x 2 2 3 6= − + = − + − + + +− +∫ ∫ ∫ -----------------------------------------------------------------------------------

-------------------------------------------

2

3

x 1I dx

(x 1)

+=

−∫ by partial fractions 3 2

1 1 1I 2 dx 2 dx dx

(x 1) (x 1) x 1= + +

− − −∫ ∫ ∫ ,

then2 1

2

2 2 1 2I (x 1) (x 1) ln(x 1) c ln(x 1) c

2 1 (x 1) x 1

− −= − − − − + − + = − − + − +− −

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2xI dx(x 1)(x 4)

= + +∫ by partial fractions

2 2

1 1 1 x 4 1I dx dx dx

5 x 1 5 x 4 5 x 4= − + +

+ + +∫ ∫ ∫

2 1

2 2

1 1 1 2x 4 1 1 1 2 xdx dx dx ln(x 1) ln(x 4) tan c

5 x 1 10 x 4 5 x 4 5 10 5 2

− = − + + = − + + + + + + + + ∫ ∫ ∫

{ }2

2 1 1

2

1 2 x 1 x 4 2 xln(x 4) 2ln(x 1) tan c ln tan c

10 5 2 10 (x 1) 5 2

− − + = + − + + + = + + + -----------------------------------------------------------------------------------

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( )

3

22

2x x 3I dx

x 1

+ +=

+∫ by partial fractions

( )22

2

2x 3 xI dx

x 1 x 1

− = + + +

∫

( ) ( )2 22

2 2

2x 1 xdx 3 dx dx

x 1 x 1 x 1= + −

+ + +∫ ∫ ∫ , now take each integrand

separately so ( )2

2

2xdx ln x 1

x 1= +

+∫ ,( )

22

xdx

x 1+∫ let

2 duu x 1 dx

2x= + ⇒ = ,

then2

1 1du2 u∫

( )

1

21 u 12 1 2 x 1

−

= = − − + ,

( )2

2

13 dx

x 1+∫ let



2x tan dx sec d= θ ⇒ = θ θ , then

( )

2 22

2 4 22

sec sec 1 33 d 3 d 3 d 3 cos d (cos2 1)d

sec sec 2tan 1

θ θθ = θ = θ = θ θ = θ + θ

θ θθ +∫ ∫ ∫ ∫ ∫

2

2

3 sin2 3 3 3 tan

(sin cos ) (tan cos )2 2 2 2 2 sec

θ θ = + θ = θ θ + θ = θ θ + θ = + θ θ

1

2

3 xtan x

2 x 1

− = + + , ( )

( )2 1

2 2

3 x 1I ln x 1 tan x c

2 x 1 2 x 1

− = + + + + + + +

( )( ) ( )

( )( )

2 1 2 1

2 2 2

3x 1 3 1 3x 3ln x 1 tan x c ln x 1 tan x c

2 22 x 1 2 x 1 2 x 1

− −+= + + + + + = + + + +

+ + +

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1 sinxI dx

sinx(1 cosx)

+=

+∫ let

1

2

x 2t tan x 2tan t dx dt

2 1 t

− = ⇒ = ⇒ = +

, then

22 2

22 2 22

22 2

2t 2t1 1

2 1 t 2t1 t 1 tI dt 2tdt 2dt2t(1 t )1 t 2t(1 t ) 2t(1 t )2t 1 t

2t11 t1 t 1 t

+ + + ++ += ⋅ = ⋅ = ⋅−+ + + − − ++ ++ +

∫ ∫ ∫

( )

2 2 2 2

2 2

t 2t 1 t 2t 1 t 2t 1 t 2t 12dt 2dt dt dt dt dt

4t 2t 2t 2t 2t2t 1 t 1 t

+ + + + + += ⋅ = ⋅ = = + +

+ + −∫ ∫ ∫ ∫ ∫ ∫ 2

21 1 1 1 t 1 1 x x 1 xtdt 1dt dt t lnt c tan tan ln tan c

2 2 t 2 2 2 4 2 2 2 2

= + + = ⋅ + + + = + + +

∫ ∫ ∫ -----------------------------------------------------------------------------------

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1I dx

5 3cosx=

−∫ let 2

x 2t tan dx dt

2 1 t

= ⇒ = + , then 2 2

2

1 2I dt

3(1 t ) 1 t5

1 t

= ⋅− +−+

∫

2 2 2 2 2 22

2 2 2 1 1 1dt dt dt dt dt

15(1 t ) 3(1 t ) 5 5t 3 3t 8t 2 4t 1 4 t4

= = = = =+ − − + − + + + +

∫ ∫ ∫ ∫ ∫

1 1 11 1 t 1 1 xtan c tan (2t) c tan 2tan c

1 14 2 2 22 2

− − −

= ⋅ + = + = + -----------------------------------------------------------------------------------

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( )3 3

1I dx

x a x=

+∫ let3

2

dtt x dx

3x= ⇒ = , then ( ) ( )3 3 3

1 1 1 1I dt dt

3 3x a t t a t= =

+ +∫ ∫

, by partial fractions3

3 3 3 3 3 3

1 1 1 1 1 xI dt dt ln c

3a t 3a a t 3a a x

= − = + + +

∫ ∫ -----------------------------------------------------------------------------------

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2I x lnxdx= ∫ letdu 1

u lnxdx x

= ⇒ = &3

2dv xx v

dx 3= ⇒ = , then by the parts

formula we have3 3 3 3

2x lnx 1 x lnx 1 x x 1I xdx c lnx c

3 3 3 3 3 3 3

= − = − ⋅ + = − + ∫

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I xarctanxdx= ∫ let 2

du 1u arctanx

dx 1 x= ⇒ =

+&

2dv xx v

dx 2= ⇒ = , then by the

parts formula2 2 2 2

2 2 2

x arctanx 1 x x arctanx 1 1 x 1I dx dx dx

2 2 1 x 2 2 1 x 1 x

+= − = − − + + +

∫ ∫ ∫

( )2 2

2

x arctanx 1 1 x arctanx 11dx dx x arctanx

2 2 1 x 2 2

= − − = − − + ∫ ∫

( )22 2 x 1x arctanx x 1 x 1 x x

arctanx arctanx arctanx c2 2 2 2 2 2 2 2

+ = − + = + − = − + -----------------------------------------------------------------------------------

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1I xsinxcosxdx xsin2xdx

2= =∫ ∫ let

duu x 1

dx= ⇒ = &

dv cos2xsin2x v

dx 2= = = − ,

then1 x 1 1 x

I cos2x cos2xdx sin2x cos2x c2 2 2 8 4

= − + = − + ∫

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2I ln xdx lnx lnxdx= = ⋅∫ ∫ let du 1u lnxdx x

= ⇒ = &

dvlnx v xlnx x x(lnx 1)

dx= ⇒ = − = − , then

( )2 2I lnx(xlnx x) (lnx 1)dx xln x 2xlnx 2x c x ln x 2lnx 2 c= − − − = − + + = − + +∫ -----------------------------------------------------------------------------------

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xI xa dx= ∫ letdu

u x 1dx

= ⇒ = &x

xdv aa v

dx lna= ⇒ = , then

x xxa a

I dxlna lna

= − ∫

x x x

xa 1 a a 1c x clna lna lna lna lna

= − ⋅ + = − + -----------------------------------------------------------------------------------

-------------------------------------------



116 4

1

0 2

xI dx

1 x

=+

∫ let 4 3x t dx 4t dt= ⇒ = , x 0 t 0= ⇒ = , x 16 t 2= ⇒ = , then

2 2 43

2 2

0 0

t tI 4t dt 4 dt

1 t 1 t

= ⋅ =+ +∫ ∫

by long division4

2

2 2

t 1t 1

1 t 1 t= + −

+ +, then we

have

22 32 1

2

0 0

1 4tI 4 t 1dt 4tan t 4t 7.1

1 t 3

− = + − = + − ≅ + ∫

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2

1I dx

1 x x=

+ +∫ let ( ) ( ) ( )

1 12 22 2

dt 1t x 1 x x 1 2x 1 1 x x

dx 2

−− = + + ⇒ − = + + + ⇒

2

2 2

dt 2x 1 2x 1 2 1 x x 2x 1 2(t x) 2(t x)1 dx dt

dx 2(t x) 2x 1 2(t x)2 1 x x 2 1 x x

+ + + + + + + − −= + = = ⇒ =

− + + −+ + + +

,

then ( ) ( )22 2I dt dt ln 2t 1 c ln 2x 1 2 1 x x c

2x 1 2t 2x 2t 1= = = + + = + + + + +

+ + − +∫ ∫ -----------------------------------------------------------------------------------

-------------------------------------------

2 2

4

a xI dx

x

−= ∫ let

2 2

2 2

1 1 dt 1t x dx xdt

x t dx x= ⇒ = ⇒ = − ⇒ = − , then it follows

that

2

22 2

2

1a

tI dt t a t 1 dt1

t

−= − = − − ⋅∫ ∫ let

2 2

2

duu a t 1 dt

2a t= − ⇒ = , therefore

( )

3 33 22 2

2 2 22 2 2 2 2

1 1 u 1 1 aI u du c a t 1 c 1 c

32a 2a 3a 3a x

2

= − ⋅ = − + = − − + = − − +

∫ , now we

finally have the following result; ( )3

2 2 2

2 2

a xI c

3a x

−= − +

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2x x

1I dxe 2e

=−∫ let

x

x

dtt e :lnt x dxe

= = ⇒ = , then 2 3 2

1 dt 1I dtt 2t t t 2t

= ⋅ =− −∫ ∫

2

1dt

t (t 2)=

−∫ now by the partial fraction theorem we have,

2

1 1 1 t 2I dt dt

4 t 2 4 t

+= −

−∫ ∫

( ) ( )x

2 x

1 1 1 1 1 1 1 1 1 1 x 1dt dt dt ln t 2 lnt c ln e 2 c

4 t 2 4 t 2 t 4 4 2t 2e 4 4= − − = − − + + = − + − +

−∫ ∫ ∫ -----------------------------------------------------------------------------------

-------------------------------------------



( )

2x

1x 4

eI dx

e 1

=+

∫ letx

x

dtt e 1 dx

e= + ⇒ = , then

2x x

1 1 1x

4 4 4

e dt e t 1I dt dt

et t t

−= ⋅ = =∫ ∫ ∫

( ) ( ) ( )3 1 7 3 3 3

x x4 4 4 4 4 44 4 4 4

t dt t dt t t c 3t 7 t c 3e 4 e 1 c

7 3 21 21

− = − = − + = − + = − + +

∫ ∫ -----------------------------------------------------------------------------------

-------------------------------------------

ln5 x x

x

0

e e 1I dx

e 3

−=

+∫ 2 x

x

2tt e 1 dx dt

e= − ⇒ = , x 0 t 0= ⇒ = , x ln5 t 2= ⇒ = , then

2 2

2

0

tI 2 dt

t 4=

+∫ let 2t 2tan dt 2sec d= θ ⇒ = θ θ , then

2 2 2

2

0

4tan 2secI 2 d

4tan 4

θ⋅ θ= θ

θ +∫

[ ]22 22 2

22 1

2 0

0 0 0

8tan sec t t2 d 4 tan d 4 tan 4 tan 4

4sec 2 2

− θ ⋅ θ = θ = θ θ = θ − θ = − = − π

θ

∫ ∫ -----------------------------------------------------------------------------------

-------------------------------------------

ln2x

0

I e 1 dx= − ⋅∫ letx

x

dtt e 1 dx

e= − ⇒ = , x 0 t 0= ⇒ = , x ln2 t 1= ⇒ = , then

1 1

x

0 0

t tI dt dt

e t 1= =

+∫ ∫ let 2 2t tan dt 2sec tan d= θ ⇒ = θ θ θ , then

1 2 2

2

0

tan secI 2 d

tan 1

θ⋅ θ= θ

θ +∫

[ ] ( )1 1

12 1

00

0

42 tan d 2 tan 2 t tan t 2 1 2

4 2 2

− π π − π = θ θ = θ − θ = − = − = − = ∫ -----------------------------------------------------------------------------------

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1 lnxI dx

x

+= ∫ let t 1 lnx dx xdt= + ⇒ = , then

( )3 3

2 22 2

I t dt t c 1 lnx c3 3

= ⋅ = + = + +∫ -----------------------------------------------------------------------------------

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( )4

0

sin cosI d

3 sin2

π

θ + θ= θ

+ θ∫ letdt

t sin cos dsin cos

= θ − θ ⇒ θ =θ + θ

, t 04

πθ = ⇒ = ,

0 t 1θ = ⇒ = − , then0

1

dtI

3 sin2−

=+ θ∫ , now consider the following

( )22t sin cos= θ − θ



( )

0 02 2 2

221 1

dt dtsin cos 2sin cos 1 t sin2 I

4 t3 1 t− −

= θ + θ − θ θ ⇒ − = θ ⇒ = =−+ −∫ ∫ , so

therefore we have that

0 1

1

1 2 t 1 1 1 1 ln3I ln ln1 ln ln

4 2 t 4 3 4 3 4

−

−

+ = = − = =

− -----------------------------------------------------------------------------------

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( ) ( ){ } ( )1sinx sinx

I tanxdx dx dx ln cosx c ln cosx c ln secx ccosx cosx

−−= = = − = − + = + = +∫ ∫ ∫

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These are a good selection of integrals, the solutions cover the main techniques that are required to attack most integration problems.

Integration e.g.

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