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

Aug 6, 2016

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

Explanation:

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

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

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

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

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

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

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

= $- 2 \ln | x | + 2 \ln | 4 x - 1 | + c$