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

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### 2 Answers

We have to find the indefinite integral of y=1/(x^2-x)

Int[ 1/(x^2-x) dx]

=> Int[ 1/x(x - 1) dx]

=> Int[ 1/(x-1) - 1/x dx]

=> Int [1/(x - 1) dx] - Int[ 1/x dx]

=> ln(x - 1) - ln x

=> ln[(x - 1)/x]

**The required integral is ln[(x - 1)/x)] + C**

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**