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

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### 1 Answer

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**