# What is the antiderivative of 1/(x^2-x)?

Oct 17, 2016

$\ln \left(\frac{x - 1}{x}\right) + C$

#### Explanation:

Transform using partial fractions
$\frac{1}{{x}^{2} - x} = \frac{1}{x \left(x - 1\right)} = \frac{A}{x} + \frac{B}{x - 1}$
$= \frac{A \left(x - 1\right) + B x}{x \left(x - 1\right)}$
Equating LHS and RHS
$1 = A \left(x - 1\right) + B x$
Firstly let x=0 => $1 = - A$, so $A = - 1$
Let x=1, then $1 = B$
So $\frac{1}{{x}^{2} - x} = \frac{1}{x - 1} - \frac{1}{x}$
So$\int \frac{\mathrm{dx}}{{x}^{2} - x} = \int \frac{\mathrm{dx}}{x - 1} - \int \frac{\mathrm{dx}}{x}$
$= \ln \left(x - 1\right) - \ln x$
$= \ln \left(\frac{x - 1}{x}\right)$
Check by differentiating $\ln \left(x - 1\right) - \ln x$
$d \frac{\ln \left(x - 1\right) - \ln x}{\mathrm{dx}} = \frac{1}{x - 1} - \frac{1}{x} = \frac{x - x + 1}{{x}^{2} - x} = \frac{1}{{x}^{2} - x}$