X20 integrals of rational functions
Transcript of X20 integrals of rational functions
![Page 1: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/1.jpg)
Integrals of Rational Functions
![Page 2: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/2.jpg)
In this section we will show how to integrate
rational functions, that is, functions of the form P(x)
Q(x)
where P and Q are polynomials.
Integrals of Rational Functions
![Page 3: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/3.jpg)
In this section we will show how to integrate
rational functions, that is, functions of the form P(x)
Q(x)
where P and Q are polynomials.
Rational Decomposition Theorem
Given reduced P/Q where deg P < deg Q,
then P/Q = F1 + F2 + .. + Fn where
Fi = or A
(ax + b)k
Ax + B(ax2 + bx + c)k
and that (ax + b)k or (ax2 + bx + c)k are
factors of Q(x).
Integrals of Rational Functions
![Page 4: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/4.jpg)
Therefore finding the integrals of rational
functions are reduced to finding integrals of
or A
(ax + b)kAx + B
(ax2 + bx + c)k
Integrals of Rational Functions
![Page 5: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/5.jpg)
Therefore finding the integrals of rational
functions are reduced to finding integrals of
or A
(ax + b)kAx + B
(ax2 + bx + c)k
Integrals of Rational Functions
The integrals ∫ are straight forward
with the substitution method by setting
u = ax + b.
dx(ax + b)k
![Page 6: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/6.jpg)
Therefore finding the integrals of rational
functions are reduced to finding integrals of
or A
(ax + b)kAx + B
(ax2 + bx + c)k
Integrals of Rational Functions
The integrals ∫ are straight forward
with the substitution method by setting
u = ax + b.
dx(ax + b)k
To integrate ∫ dx, we need to
complete the square for the denominator.
Ax + B(ax2 + bx + c)k
![Page 7: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/7.jpg)
xx2 + 6x + 10 Example: Find ∫ dx
Integrals of Rational Functions
![Page 8: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/8.jpg)
x2 + 6x + 10 - its irreducible.Complete the square on
Integrals of Rational Functionsx
x2 + 6x + 10 Example: Find ∫ dx
![Page 9: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/9.jpg)
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x ) + 10
Integrals of Rational Functionsx
x2 + 6x + 10 Example: Find ∫ dx
![Page 10: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/10.jpg)
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
Integrals of Rational Functionsx
x2 + 6x + 10 Example: Find ∫ dx
![Page 11: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/11.jpg)
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
Integrals of Rational Functionsx
x2 + 6x + 10 Example: Find ∫ dx
![Page 12: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/12.jpg)
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
xx2 + 6x + 10 ∫ dx
= ∫x
(x + 3)2 + 1 dx
Hence
Integrals of Rational Functionsx
x2 + 6x + 10 Example: Find ∫ dx
![Page 13: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/13.jpg)
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
xx2 + 6x + 10 ∫ dx
Set u = x + 3
= ∫x
(x + 3)2 + 1 dx
substitutionHence
Integrals of Rational Functionsx
x2 + 6x + 10 Example: Find ∫ dx
![Page 14: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/14.jpg)
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
xx2 + 6x + 10 ∫ dx
Set u = x + 3
x = u – 3
dx = du= ∫
x(x + 3)2 + 1
dx
substitutionHence
Integrals of Rational Functionsx
x2 + 6x + 10 Example: Find ∫ dx
![Page 15: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/15.jpg)
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
xx2 + 6x + 10 ∫ dx
Set u = x + 3
x = u – 3
dx = du= ∫
x(x + 3)2 + 1
dx
substitution
= ∫u – 3 u2 + 1
du
Hence
Integrals of Rational Functionsx
x2 + 6x + 10 Example: Find ∫ dx
![Page 16: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/16.jpg)
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
xx2 + 6x + 10 ∫ dx
Set u = x + 3
x = u – 3
dx = du= ∫
x(x + 3)2 + 1
dx
substitution
= ∫u – 3 u2 + 1
du
= ∫u
u2 + 1 du – 3 ∫1
u2 + 1 du
Hence
Integrals of Rational Functionsx
x2 + 6x + 10 Example: Find ∫ dx
![Page 17: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/17.jpg)
x2 + 6x + 10 - its irreducible.Complete the square on
x2 + 6x + 10 = (x2 + 6x + 9) + 10 – 9
= (x + 3)2 + 1
xx2 + 6x + 10 ∫ dx
Set u = x + 3
x = u – 3
dx = du= ∫
x(x + 3)2 + 1
dx
substitution
= ∫u – 3 u2 + 1
du
= ∫u
u2 + 1 du – 3 ∫1
u2 + 1 du
Hence
Integrals of Rational Functions
Set w = u2 + 1
substitution
xx2 + 6x + 10 Example: Find ∫ dx
![Page 18: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/18.jpg)
Set w = u2 + 1
= 2u
substitution
dwdu
Integrals of Rational Functions
![Page 19: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/19.jpg)
Set w = u2 + 1
= 2u
substitution
dwdu
du =dw2u
Integrals of Rational Functions
![Page 20: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/20.jpg)
Set w = u2 + 1
= 2u
substitution
dwdu
du =dw2u
= ∫u w
– 3 ∫1
u2 + 1 dudw2u
Integrals of Rational Functions
![Page 21: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/21.jpg)
Set w = u2 + 1
= 2u
substitution
dwdu
du =dw2u
= ∫u w
– 3 ∫1
u2 + 1 dudw2u
= ½ ∫1 w – 3 tan-1(u) + c dw
Integrals of Rational Functions
![Page 22: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/22.jpg)
Set w = u2 + 1
= 2u
substitution
dwdu
du =dw2u
= ∫u w
– 3 ∫1
u2 + 1 dudw2u
= ½ ∫1 w – 3 tan-1(u) + c dw
= ½ Ln(w) – 3 tan-1(x + 3) + c
Integrals of Rational Functions
![Page 23: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/23.jpg)
Set w = u2 + 1
= 2u
substitution
dwdu
du =dw2u
= ∫u w
– 3 ∫1
u2 + 1 dudw2u
= ½ ∫1 w – 3 tan-1(u) + c dw
= ½ Ln(w) – 3 tan-1(x + 3) + c
= ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c
Integrals of Rational Functions
![Page 24: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/24.jpg)
Set w = u2 + 1
= 2u
substitution
dwdu
du =dw2u
= ∫u w
– 3 ∫1
u2 + 1 dudw2u
= ½ ∫1 w – 3 tan-1(u) + c dw
= ½ Ln(w) – 3 tan-1(x + 3) + c
= ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c
Integrals of Rational Functions
3xx2 + 7x + 10 Example: Find ∫ dx
![Page 25: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/25.jpg)
Set w = u2 + 1
= 2u
substitution
dwdu
du =dw2u
= ∫u w
– 3 ∫1
u2 + 1 dudw2u
= ½ ∫1 w – 3 tan-1(u) + c dw
= ½ Ln(w) – 3 tan-1(x + 3) + c
= ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c
Integrals of Rational Functions
Since x2 + 7x + 10 = (x + 2)(x + 5), we can
decompose the rational expression.
3xx2 + 7x + 10 Example: Find ∫ dx
![Page 26: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/26.jpg)
Set w = u2 + 1
= 2u
substitution
dwdu
du =dw2u
= ∫u w
– 3 ∫1
u2 + 1 dudw2u
= ½ ∫1 w – 3 tan-1(u) + c dw
= ½ Ln(w) – 3 tan-1(x + 3) + c
= ½ Ln((x + 3)2 + 1) – 3 tan-1(x + 3) + c
Integrals of Rational Functions
3xx2 + 7x + 10 Example: Find ∫ dx
Since x2 + 7x + 10 = (x + 2)(x + 5), we can
decompose the rational expression.
Specifically 3x(x + 2)(x + 5) =
A(x + 2)
+B
(x + 5)
![Page 27: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/27.jpg)
Integrals of Rational Functions
Clear denominators: 3x(x + 2)(x + 5) =
A(x + 2)
+B
(x + 5)
![Page 28: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/28.jpg)
Integrals of Rational Functions
3x = A(x + 5) + B(x + 2)
Clear denominators: 3x(x + 2)(x + 5) =
A(x + 2)
+B
(x + 5)
![Page 29: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/29.jpg)
Integrals of Rational Functions
3x = A(x + 5) + B(x + 2)
Evaluate at x = -5, we get -15 = -3B
Clear denominators: 3x(x + 2)(x + 5) =
A(x + 2)
+B
(x + 5)
![Page 30: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/30.jpg)
Integrals of Rational Functions
3x = A(x + 5) + B(x + 2)
Evaluate at x = -5, we get -15 = -3B so B = 5
Clear denominators: 3x(x + 2)(x + 5) =
A(x + 2)
+B
(x + 5)
![Page 31: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/31.jpg)
Integrals of Rational Functions
3x = A(x + 5) + B(x + 2)
Evaluate at x = -5, we get -15 = -3B so B = 5
Evaluate at x = -2, we get -6 = 3A
Clear denominators: 3x(x + 2)(x + 5) =
A(x + 2)
+B
(x + 5)
![Page 32: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/32.jpg)
Integrals of Rational Functions
3x = A(x + 5) + B(x + 2)
Evaluate at x = -5, we get -15 = -3B so B = 5
Evaluate at x = -2, we get -6 = 3A so A = -2
Clear denominators: 3x(x + 2)(x + 5) =
A(x + 2)
+B
(x + 5)
![Page 33: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/33.jpg)
Integrals of Rational Functions
Hencex
(x + 2)(x + 5) =-2
(x + 2) +
5(x + 5)
3x = A(x + 5) + B(x + 2)
Evaluate at x = -5, we get -15 = -3B so B = 5
Evaluate at x = -2, we get -6 = 3A so A = -2
Clear denominators: 3x(x + 2)(x + 5) =
A(x + 2)
+B
(x + 5)
![Page 34: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/34.jpg)
Integrals of Rational Functions
Hencex
(x + 2)(x + 5) =-2
(x + 2) +
5(x + 5)
3x = A(x + 5) + B(x + 2)
Evaluate at x = -5, we get -15 = -3B so B = 5
Evaluate at x = -2, we get -6 = 3A so A = -2
3xx2 + 7x + 10 So ∫ dx
Clear denominators: 3x(x + 2)(x + 5) =
A(x + 2)
+B
(x + 5)
![Page 35: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/35.jpg)
Integrals of Rational Functions
Hencex
(x + 2)(x + 5) =-2
(x + 2) +
5(x + 5)
3x = A(x + 5) + B(x + 2)
Evaluate at x = -5, we get -15 = -3B so B = 5
Evaluate at x = -2, we get -6 = 3A so A = -2
3xx2 + 7x + 10 So ∫ dx
= ∫ dx -2
(x + 2) +
5(x + 5)
Clear denominators: 3x(x + 2)(x + 5) =
A(x + 2)
+B
(x + 5)
![Page 36: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/36.jpg)
Integrals of Rational Functions
Hencex
(x + 2)(x + 5) =-2
(x + 2) +
5(x + 5)
3x = A(x + 5) + B(x + 2)
Evaluate at x = -5, we get -15 = -3B so B = 5
Evaluate at x = -2, we get -6 = 3A so A = -2
3xx2 + 7x + 10 So ∫ dx
= -2Ln(x + 2) + 5Ln(x + 5) + c
= ∫ dx -2
(x + 2) +
5(x + 5)
Clear denominators: 3x(x + 2)(x + 5) =
A(x + 2)
+B
(x + 5)
![Page 37: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/37.jpg)
Integrals of Rational Functions
1(x2 + 1)2x Example: Find ∫ dx
![Page 38: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/38.jpg)
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+Cx + D(x2 + 1)2 +
Ex
1(x2 + 1)2x Example: Find ∫ dx
![Page 39: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/39.jpg)
Integrals of Rational Functions
1(x2 + 1)2x Example: Find ∫ dx
1(x2 + 1)2x =
Ax + B(x2 + 1)
+Cx + D(x2 + 1)2 +
Ex
Clear denominators
![Page 40: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/40.jpg)
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+Cx + D(x2 + 1)2 +
Ex
Clear denominators
1 = (Ax + B)(x2 + 1) x + (Cx + D) x +E (x2 + 1)2
1(x2 + 1)2x Example: Find ∫ dx
![Page 41: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/41.jpg)
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+Cx + D(x2 + 1)2 +
Ex
Clear denominators
1 = (Ax + B)(x2 + 1) x + (Cx + D) x +E (x2 + 1)2
Evaluate this at x = 0, we get 1 = E
1(x2 + 1)2x Example: Find ∫ dx
![Page 42: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/42.jpg)
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+Cx + D(x2 + 1)2 +
Ex
Clear denominators
1 = (Ax + B)(x2 + 1) x + (Cx + D) x +E (x2 + 1)2
Evaluate this at x = 0, we get 1 = E
Hence 1 = (Ax + B)(x2 + 1) x + (Cx + D)x +1(x2 + 1)2
1(x2 + 1)2x Example: Find ∫ dx
![Page 43: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/43.jpg)
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+Cx + D(x2 + 1)2 +
Ex
Clear denominators
1 = (Ax + B)(x2 + 1) x + (Cx + D) x +E (x2 + 1)2
Evaluate this at x = 0, we get 1 = E
Hence 1 = (Ax + B)(x2 + 1) x + (Cx + D)x +1(x2 + 1)2
There is no x4-term,
1(x2 + 1)2x Example: Find ∫ dx
![Page 44: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/44.jpg)
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+Cx + D(x2 + 1)2 +
Ex
Clear denominators
1 = (Ax + B)(x2 + 1) x + (Cx + D) x +E (x2 + 1)2
Evaluate this at x = 0, we get 1 = E
Hence 1 = (Ax + B)(x2 + 1) x + (Cx + D)x +1(x2 + 1)2
There is no x4-term, hence Ax4 + x4 = 0,
1(x2 + 1)2x Example: Find ∫ dx
![Page 45: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/45.jpg)
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+Cx + D(x2 + 1)2 +
Ex
Clear denominators
1 = (Ax + B)(x2 + 1) x + (Cx + D) x +E (x2 + 1)2
Evaluate this at x = 0, we get 1 = E
Hence 1 = (Ax + B)(x2 + 1) x + (Cx + D)x +1(x2 + 1)2
There is no x4-term, hence Ax4 + x4 = 0, so A = -1
1(x2 + 1)2x Example: Find ∫ dx
![Page 46: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/46.jpg)
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+Cx + D(x2 + 1)2 +
Ex
Clear denominators
1 = (Ax + B)(x2 + 1) x + (Cx + D) x +E (x2 + 1)2
Evaluate this at x = 0, we get 1 = E
Hence 1 = (Ax + B)(x2 + 1) x + (Cx + D)x +1(x2 + 1)2
There is no x4-term, hence Ax4 + x4 = 0, so A = -1
There is no x3-term,
1(x2 + 1)2x Example: Find ∫ dx
![Page 47: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/47.jpg)
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+Cx + D(x2 + 1)2 +
Ex
Clear denominators
1 = (Ax + B)(x2 + 1) x + (Cx + D) x +E (x2 + 1)2
Evaluate this at x = 0, we get 1 = E
Hence 1 = (Ax + B)(x2 + 1) x + (Cx + D)x +1(x2 + 1)2
There is no x4-term, hence Ax4 + x4 = 0, so A = -1
There is no x3-term, hence Bx3 = 0,
1(x2 + 1)2x Example: Find ∫ dx
![Page 48: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/48.jpg)
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+Cx + D(x2 + 1)2 +
Ex
Clear denominators
1 = (Ax + B)(x2 + 1) x + (Cx + D) x +E (x2 + 1)2
Evaluate this at x = 0, we get 1 = E
Hence 1 = (Ax + B)(x2 + 1) x + (Cx + D)x +1(x2 + 1)2
There is no x4-term, hence Ax4 + x4 = 0, so A = -1
There is no x3-term, hence Bx3 = 0, so B = 0
1(x2 + 1)2x Example: Find ∫ dx
![Page 49: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/49.jpg)
Integrals of Rational Functions
1(x2 + 1)2x =
Ax + B(x2 + 1)
+Cx + D(x2 + 1)2 +
Ex
Clear denominators
1 = (Ax + B)(x2 + 1) x + (Cx + D) x +E (x2 + 1)2
Evaluate this at x = 0, we get 1 = E
Hence 1 = (Ax + B)(x2 + 1) x + (Cx + D)x +1(x2 + 1)2
There is no x4-term, hence Ax4 + x4 = 0, so A = -1
There is no x3-term, hence Bx3 = 0, so B = 0
So we've 1 = -x (x2 + 1) x + (Cx + D)x +1(x2 + 1)2
1(x2 + 1)2x Example: Find ∫ dx
![Page 50: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/50.jpg)
Integrals of Rational Functions
There is no x-term in
1 = -x (x2 + 1) x + (Cx + D)x +1(x2 + 1)2
![Page 51: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/51.jpg)
Integrals of Rational Functions
There is no x-term in
1 = -x (x2 + 1) x + (Cx + D)x +1(x2 + 1)2
Hence Dx = 0, or D = 0.
![Page 52: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/52.jpg)
Integrals of Rational Functions
There is no x-term in
1 = -x (x2 + 1) x + (Cx + D)x +1(x2 + 1)2
Hence Dx = 0, or D = 0. So the expression is
1 = -x2 (x2 + 1) + Cx2 +1(x2 + 1)2
![Page 53: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/53.jpg)
Integrals of Rational Functions
There is no x-term in
1 = -x (x2 + 1) x + (Cx + D)x +1(x2 + 1)2
Hence Dx = 0, or D = 0. So the expression is
1 = -x2 (x2 + 1) + Cx2 +1(x2 + 1)2
Expand this
![Page 54: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/54.jpg)
Integrals of Rational Functions
There is no x-term in
1 = -x (x2 + 1) x + (Cx + D)x +1(x2 + 1)2
Hence Dx = 0, or D = 0. So the expression is
1 = -x2 (x2 + 1) + Cx2 +1(x2 + 1)2
Expand this
1 = -x4 – x2 Cx2 + x4 + 2x2 + 1+
![Page 55: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/55.jpg)
Integrals of Rational Functions
There is no x-term in
1 = -x (x2 + 1) x + (Cx + D)x +1(x2 + 1)2
Hence Dx = 0, or D = 0. So the expression is
1 = -x2 (x2 + 1) + Cx2 +1(x2 + 1)2
Expand this
1 = -x4 – x2 Cx2 + x4 + 2x2 + 1+
Hence Cx2 + x2 = 0 or C = -1
![Page 56: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/56.jpg)
Integrals of Rational Functions
There is no x-term in
1 = -x (x2 + 1) x + (Cx + D)x +1(x2 + 1)2
Hence Dx = 0, or D = 0. So the expression is
1 = -x2 (x2 + 1) + Cx2 +1(x2 + 1)2
Expand this
1 = -x4 – x2 Cx2 + x4 + 2x2 + 1+
Hence Cx2 + x2 = 0 or C = -1
Put it all together
1(x2 + 1)2x =
-x(x2 + 1)
+-x
(x2 + 1)2 +1x
![Page 57: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/57.jpg)
Integrals of Rational Functions
1(x2 + 1)2x Therefore ∫ dx
= ∫-x dx
(x2 + 1) +
-x dx (x2 + 1)2 +
dxx ∫ ∫
![Page 58: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/58.jpg)
Integrals of Rational Functions
1(x2 + 1)2x Therefore ∫ dx
= ∫-x dx
(x2 + 1) +
-x dx (x2 + 1)2 +
dxx ∫ ∫
substitution
set u = x2 + 1
![Page 59: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/59.jpg)
Integrals of Rational Functions
1(x2 + 1)2x Therefore ∫ dx
= ∫-x dx
(x2 + 1) +
-x dx (x2 + 1)2 +
dxx ∫ ∫
substitution
set u = x2 + 1 dudx = 2x
![Page 60: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/60.jpg)
Integrals of Rational Functions
1(x2 + 1)2x Therefore ∫ dx
= ∫-x dx
(x2 + 1) +
-x dx (x2 + 1)2 +
dxx ∫ ∫
substitution
set u = x2 + 1 dudx = 2x
du2x
dx =
![Page 61: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/61.jpg)
Integrals of Rational Functions
1(x2 + 1)2x Therefore ∫ dx
= ∫-x dx
(x2 + 1) +
-x dx (x2 + 1)2 +
dxx ∫ ∫
substitution
set u = x2 + 1 dudx = 2x
du2x
dx = = ∫-x u
du2x
+ ∫-x u2
du2x
+ Ln(x)
![Page 62: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/62.jpg)
Integrals of Rational Functions
1(x2 + 1)2x Therefore ∫ dx
= ∫-x dx
(x2 + 1) +
-x dx (x2 + 1)2 +
dxx ∫ ∫
substitution
set u = x2 + 1 dudx = 2x
du2x
dx = = ∫-x u
du2x
+ ∫-x u2
du2x
+ Ln(x)
= - ½ ∫duu + Ln(x)½ ∫
duu2–
![Page 63: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/63.jpg)
Integrals of Rational Functions
1(x2 + 1)2x Therefore ∫ dx
= ∫-x dx
(x2 + 1) +
-x dx (x2 + 1)2 +
dxx ∫ ∫
substitution
set u = x2 + 1 dudx = 2x
du2x
dx = = ∫-x u
du2x
+ ∫-x u2
du2x
+ Ln(x)
= - ½ ∫duu + Ln(x)½ ∫
duu2–
= - ½ Ln(u) + ½ u-1 + Ln(x) + c
![Page 64: X20 integrals of rational functions](https://reader034.fdocuments.net/reader034/viewer/2022051504/58f0e2f41a28ab953b8b45fd/html5/thumbnails/64.jpg)
Integrals of Rational Functions
1(x2 + 1)2x Therefore ∫ dx
= ∫-x dx
(x2 + 1) +
-x dx (x2 + 1)2 +
dxx ∫ ∫
substitution
set u = x2 + 1 dudx = 2x
du2x
dx = = ∫-x u
du2x
+ ∫-x u2
du2x
+ Ln(x)
= - ½ ∫duu + Ln(x)½ ∫
duu2–
= - ½ Ln(u) + ½ u-1 + Ln(x) + c
= - ½ Ln(x2 + 1) + + Ln(x) + c1
2(x2 + 1)