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

1 Answer
Oct 17, 2016

#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)#