# How do you integrate int (x^2 - 4) / (x -1) dx using partial fractions?

$\int \frac{{x}^{2} - 4}{x - 1} \mathrm{dx} = \int \left(x + 1 - \frac{3}{x - 1}\right) \mathrm{dx}$
$= \frac{1}{2} {x}^{2} + x - 3 \ln \left\mid x - 1 \right\mid + C$