# Calculus AB

I had to integrate [(x^2+1)/(x^2-x)]dx with partial fractions. My answer was 2ln abs(x-1) -ln abs(x)+C. But the answer on the answer sheet has an extra +x that I did not account for. Is that a typo or did I integrate incorrectly?

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

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