Question #59893

1 Answer
Jul 25, 2017

=c_0 x + (2 c_2 - c_3)/(x+2) + c_1 Log(x-1) + c_2 Log(x+2)+C=c0x+2c2c3x+2+c1log(x1)+c2log(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)=0c0+c1x1+c2x+c3(x+2)2x3(x1)(x+2)2=0

Solving for c_0,c_1,c_2,c_3c0,c1,c2,c3 we have

c_0=2,c_1=2/9,c_2=-56/9,c_3=-64/9c0=2,c1=29,c2=569,c3=649 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=x3(x1)(x+2)2dx=(c0+c1x1+c2x+c3(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=c0x+2c2c3x+2+c1log(x1)+c2log(x+2)+C