Compute the partial fractions:
(x^2 + 2x)/(x^2 + 1)^2 = (Ax + B)/(x^2 + 1)^2 + C/(x^2 + 1)x2+2x(x2+1)2=Ax+B(x2+1)2+Cx2+1
x^2 + 2x = Ax + B + C(x^2 + 1)x2+2x=Ax+B+C(x2+1)
Let x = 0x=0:
B + C = 0B+C=0
Let x = 1:
A + B + 2C = 3A+B+2C=3
Let x = -1:
-A + B + 2C = -1−A+B+2C=−1
Write the equation, A + B + 2C = 3A+B+2C=3, in the first row of an augmented matrix:
[
(1, 1 , 2 , | , 3)
]
Add the row for B + C = 0:
[
(1, 1 , 2 , | , 3),
(0, 1 , 1 , | , 0)
]
Add the row for -A + B + 2C = -1:
[
(1, 1 , 2 , | , 3),
(0, 1 , 1 , | , 0),
(-1, 1 , 2 , | , -1)
]
Add row 1 to row 3:
[
(1, 1 , 2 , | , 3),
(0, 1 , 1 , | , 0),
(0, 2 , 4 , | , 2)
]
Divide row 3 by 2:
[
(1, 1 , 2 , | , 3),
(0, 1 , 1 , | , 0),
(0, 1 , 2 , | , 1)
]
Subtract row 2 from row 3:
[
(1, 1 , 2 , | , 3),
(0, 1 , 1 , | , 0),
(0, 0 , 1 , | , 1)
]
Subtract row 3 from row 2:
[
(1, 1 , 2 , | , 3),
(0, 1 , 0 , | , -1),
(0, 0 , 1 , | , 1)
]
Subtract row 2 from row 1:
[
(1, 0 , 2 , | , 4),
(0, 1 , 0 , | , -1),
(0, 0 , 1 , | , 1)
]
Multiply row 3 by -2 and add to row 1:
[
(1, 0 , 0 , | , 2),
(0, 1 , 0 , | , -1),
(0, 0 , 1 , | , 1)
]
A = 2, B = -1, C = 1
(x^2 + 2x)/(x^2 + 1)^2 = (2x)/(x^2 + 1)^2 - 1/(x^2 + 1)^2 + 1/(x^2 + 1)
int(x^2 + 2x)/(x^2 + 1)^2dx = int(2x)/(x^2 + 1)^2dx - int1/(x^2 + 1)^2dx + int1/(x^2 + 1)dx
let u = x² + 1, du = 2xdx, intu^(-2)du = -u^-1
int(x^2 + 2x)/(x^2 + 1)^2dx = -1/(x^2 + 1) - int1/(x^2 + 1)^2dx + int1/(x^2 + 1)dx
The third integral is our old friend the inverse tangent:
int(x^2 + 2x)/(x^2 + 1)^2dx = -1/(x^2 + 1) - int1/(x^2 + 1)^2dx + tan^-1(x)
For the middle term I used wolframalpha
int(x^2 + 2x)/(x^2 + 1)^2dx = -1/(x^2 + 1) - 1/2(x/(x^2 + 1) + tan^-1(x)) + tan^-1(x) + C
int(x^2 + 2x)/(x^2 + 1)^2dx = - 1/2((x+ 2)/(x^2 + 1) + tan^-1(x)) + tan^-1(x) + C
int(x^2 + 2x)/(x^2 + 1)^2dx = - 1/2((x+ 2)/(x^2 + 1) - tan^-1(x)) + C
