Show how to evaluate the indefinite integral of function y=1/(x^2-x)?

We'll use partial fraction decomposition to evaluate the
given integral.

We'll first factorize the
denominator:

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

Now, we'll re-write the fraction 1/ ( x^2 - x ) as
an algebraic sum of irreducible fractions.

1/(x^2 - x) =
A/x + B/(x-1)

1 = A(x-1) +
Bx

1 = Ax - A + Bx

1 = x(A+B)
- A

A+B = 0 =>
A=-B

-A=1 => A=-1
=>B=1

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

Now, we'll evaluate the
integral:

`int` dx/ ( x^2 - x ) = -`int` dx/x + `int`
dx/(x-1)

`int` dx/ ( x^2 - x ) = -ln|x| + ln|x-1| +
C

We'll use the quotient property of
logarithms:

`int` dx/ ( x^2 - x ) = ln|(x-1)/x| +
C

The requested indefinite integral is: `int`
dx/ ( x^2 - x ) = ln|(x-1)/x| + C