Compute the partial fractions:
#(x^2 + 2x)/(x^2 + 1)^2 = (Ax + B)/(x^2 + 1)^2 + C/(x^2 + 1)#
#x^2 + 2x = Ax + B + C(x^2 + 1)#
Let #x = 0#:
#B + C = 0#
Let x = 1:
#A + B + 2C = 3#
Let x = -1:
#-A + B + 2C = -1#
Write the equation, #A + 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#
