Question

Question #59893

Answer 1

`=c_0 x + (2 c_2 - c_3)/(x+2) + c_1 Log(x-1) + c_2 Log(x+2)+C`

Explanation:

`c_0+c_1/(x - 1) + (c_2 x + c_3)/(x + 2)^2 - x^3/((x - 1)(x + 2)^2)=0`

Solving for `c_0,c_1,c_2,c_3` we have

`c_0=2,c_1=2/9,c_2=-56/9,c_3=-64/9` and then

`int x^3/((x - 1)(x + 2)^2)dx=int(c_0+c_1/(x - 1) + (c_2 x + c_3)/(x + 2)^2)dx=`

`=c_0 x + (2 c_2 - c_3)/(x+2) + c_1 Log(x-1) + c_2 Log(x+2)+C`