Question

How do you find `int (1-x)/(x(x+1)^2)dx` using partial fractions?

Answer 1

We need to find constants A,B,C such as

`(1-x)/(x(x+1)^2)=A/x+B/(x+1)+C/(x+1)^2`

Giving `x` various values we get that

`(1-x)/(x(x+1)^2)=1/x-1/(x+1)-2/(x+1)^2`

Hence

#int (1-x)/(x(x+1)^2)dx=int 1/xdx-int 1/(x+1)dx-int 2/(x+1)^2dx= lnabsx-lnabs(x+1)+2/(x+1)+c#