Question

How do you integrate `(x^3+2)/(x^2-x)` using partial fractions?

Answer 1

`=x^2/2+x-2lnx+3ln(x-1)+C`

Explanation:

First make a long division to determine
`(x^3+2)/(x^2-x)=x+1+(x+2)/(x^2-x)`
To simplify the fraction we use partial fraction
`(x+2)/(x^2-x)=(x+2)/(x(x-1))=A/x+B/(x-1)=(A(x-1)+Bx)/(x(x-1))`
So we determine A and B
`x+2=A(x-1)+Bx`
Cmparing the coefficients, we find
`1=A+B` and `2=-A` and we conclude `B=3`
`int((x^3+2)dx)/(x^2-x)=int(x+1)dx-int2dx/x+int3dx/(x-1)`
`=x^2/2+x-2lnx+3ln(x-1)+C`