Question

What is the antiderivative of `1/(x^2-x)`?

Answer 1

`ln((x-1)/x)+C`

Explanation:

Transform using partial fractions
`1/(x^2-x)=1/(x(x-1))=A/x+B/(x-1)`
`=(A(x-1)+Bx)/(x(x-1))`
Equating LHS and RHS
`1=A(x-1)+Bx`
Firstly let x=0 => `1=-A`, so `A=-1`
Let x=1, then `1=B`
So `1/(x^2-x)=1/(x-1)-1/x`
So`intdx/(x^2-x)=intdx/(x-1)-intdx/x`
`=ln(x-1)-lnx`
`=ln((x-1)/x)`
Check by differentiating `ln(x-1)-lnx`
`d(ln(x-1)-lnx)/dx=1/(x-1)-1/x=(x-x+1)/(x^2-x)=1/(x^2-x)`