# How do you integrate int (-2x-3)/(x^2-x)  using partial fractions?

Nov 17, 2016

The answer is $= 3 \ln x - 5 \ln \left(x - 1\right) + C$

#### Explanation:

Let's do the decomposition in partial fractions

The denominator $= {x}^{2} - x = x \left(x - 1\right)$

$\frac{- 2 x - 3}{x \left(x - 1\right)} = \frac{A}{x} + \frac{B}{x - 1}$

$= \frac{A \left(x - 1\right) + B x}{x \left(x - 1\right)}$

Therefore,
$- 2 x - 3 = A \left(x - 1\right) + B x$

let $x = 1$, $\implies$ $- 5 = B$

let $x = 0$, $\implies$ $- 3 = - A$ $\implies$ $A = 3$

So, $\frac{- 2 x - 3}{x \left(x - 1\right)} = \frac{3}{x} - \frac{5}{x - 1}$

$\int \frac{\left(- 2 x - 3\right) \mathrm{dx}}{x \left(x - 1\right)} = \int \frac{3 \mathrm{dx}}{x} - \int \frac{5 \mathrm{dx}}{x - 1}$

$= 3 \ln x - 5 \ln \left(x - 1\right) + C$

as $\int \frac{\mathrm{dx}}{x} = \ln x + c$