We have to write the function as a sum of simple quotients:
1/(x^2-x)= 1/x(x-1) = (A/x) + [B/(x-1)]
Bringing the 2 ratio to the same denominator, we'll have:
1 = A(x-1) + Bx
1 = Ax - A + Bx
1 = x(A+B) - A
The corresponding coefficients from the expressions from both sides of the equality, have to be equal.
So, the coefficient of x, from the left side expression, is 0, so that:
-A=1, so A=-1
So, -1+B=0, B=1.
1/x(x-1) = (-1/x) + [1/(x-1)]
Integral [1/x(x-1)]dx=Integral (-1/x)dx+Integral [1/(x-1)]dx
Integral (-1/x)dx = -ln x + C
Integral [1/(x-1)]dx = ln(x-1) + C
Integral [1/x(x-1)]dx = -ln x + ln(x-1) + C
Integral [1/x(x+1)]dx = ln [(x-1)/x]+ C