Evaluate the indefinite integral Integral [dx/(x^2-x)]

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+B=0

-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