# How do you integrate (2x-1)/(x^2-x-6) using partial fractions?

Oct 29, 2016

THe answer is $= \ln \left(x + 2\right) + \ln \left(x - 3\right) + C$

#### Explanation:

Let's factorise the denominator
$\frac{2 x - 1}{{x}^{2} - x - 6} = \frac{2 x - 1}{\left(x + 2\right) \left(x - 3\right)}$
Then we can decompose into partial fractions
Let $\frac{2 x - 1}{\left(x + 2\right) \left(x - 3\right)} = \frac{A}{x + 2} + \frac{B}{x - 3}$
$= \frac{A \left(x - 3\right) + B \left(x + 2\right)}{\left(x + 2\right) \left(x - 3\right)}$
so we can compare the numerators
$2 x - 1 = A \left(x - 3\right) + B \left(x + 2\right)$
let $x = 3$$\implies$$5 = 5 B$$\implies$$B = 1$
let $x = - 2$$\implies$$- 5 = - 5 A$$\implies$$A = 1$
and finally we have
$\frac{2 x - 1}{\left(x + 2\right) \left(x - 3\right)} = \frac{1}{x + 2} + \frac{1}{x - 3}$
Now we can integrate
$\int \frac{\left(2 x - 1\right) \mathrm{dx}}{\left(x + 2\right) \left(x - 3\right)} = \int \frac{\mathrm{dx}}{x + 2} + \int \frac{\mathrm{dx}}{x - 3}$
$= \ln \left(x + 2\right) + \ln \left(x - 3\right) + C$