# How do you use partial fractions to find the integral int (2x-3)/(x-1)^2dx?

Dec 9, 2016

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

#### Explanation:

As the denominator of the integrand is already factorized we can quickly decompose it using partial fractions:

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

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

Hence:

$A = 2$
$B = - 1$

and:

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

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