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

Jun 17, 2016

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

#### Explanation:

Partial fractions of $\frac{2}{x \left(4 x - 1\right)}$ are given by

$\frac{2}{x \left(4 x - 1\right)} \Leftrightarrow \frac{A}{x} + \frac{B}{4 x - 1}$ or

$\frac{2}{x \left(4 x - 1\right)} \Leftrightarrow \frac{A \left(4 x - 1\right) + B x}{x \left(4 x - 1\right)}$ or

$\frac{2}{x \left(4 x - 1\right)} \Leftrightarrow \frac{x \left(4 A + B\right) - A}{x \left(4 x - 1\right)}$

i.e. $A = - 2$ and $4 A + B = 0$ i.e. $B = - 4 A = 8$

Hence $\frac{2}{x \left(4 x - 1\right)} = - \frac{2}{x} + \frac{8}{4 x - 1}$ and

$\int \frac{2}{x \left(4 x - 1\right)} \mathrm{dx} = \int \left(- \frac{2}{x} + \frac{8}{4 x - 1}\right) \mathrm{dx}$

= $- 2 \ln x + 8 \times \frac{1}{4} \times \ln \left(4 x - 1\right)$

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