partial fractions and integration???

Question

~ denotes integration symbol
i need to use partial fractions to show that

~ x^2 +1 / x^2 - x .dx = x = ln ((x-1)^2/ x inside straight brackets) +c
any help would be great

Answer 1

(x^2 +1) /( x^2 - x) =
(x^2 +1) /(x( x - 1)) = -1/x + 2/(x - 1)
∫[ - 1/x + 2/(x - 1)]dx =
- ln(x) + 2ln(x - 1) + c =
ln[(x - 1)^2/x] + c

Answer 2

♠(xx+1)/(xx-x) = 1 +(1+x)/(xx-x) = 1+a/x +b/(x-1); a and b to be found;
♣ wrapping back: (1+x)/(xx-x) = (ax-a+bx)/ (xx-x), hence a=-1, a+b=1, b=2;
♦ y= 1-1/x +1/(x-1), hence Y=x +ln((C/x)*(x-1)^2); respond!