# How do you integrate int (2x-2)/((x-4)(x-1)(x-6))  using partial fractions?

Nov 15, 2016

The answer is $= - \ln \left(x - 4\right) + \ln \left(x - 6\right) + C$

#### Explanation:

Let's start the decomposition into partial fractions

$\frac{2 x - 2}{\left(x - 4\right) \left(x - 1\right) \left(x - 6\right)} = \frac{2 \cancel{x - 1}}{\left(x - 4\right) \cancel{x - 1} \left(x - 6\right)}$

$= \frac{2}{\left(x - 4\right) \left(x - 6\right)} = \frac{A}{x - 4} + \frac{B}{x - 6}$

$= \frac{A \left(x - 6\right) + B \left(x - 4\right)}{\left(x - 4\right) \left(x - 6\right)}$

So, $2 = \left(A \left(x - 6\right) + B \left(x - 4\right)\right)$

Let $x = 4$ $\implies$$2 = - 2 A$ $\implies$$A = - 1$

Let $x = 6$ $\implies$ $2 = 2 B$ $\implies$ $B = 1$

$\frac{2}{\left(x - 4\right) \left(x - 6\right)} = - \frac{1}{x - 4} + \frac{1}{x - 6}$

$\int \frac{2 \mathrm{dx}}{\left(x - 4\right) \left(x - 6\right)} = \int \frac{- 1 \mathrm{dx}}{x - 4} + \int \frac{1 \mathrm{dx}}{x - 6}$

$= - \ln \left(x - 4\right) + \ln \left(x - 6\right) + C$