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

Jan 25, 2017

The answer is $= - \ln \left(| 1 - 2 x |\right) + \ln \left(| 1 - x |\right) + C$

#### Explanation:

Let's start the decomposition into partial fractions

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

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

The denominators are the same ; so, we compare the numerators

$1 = A \left(1 - x\right) + B \left(1 - 2 x\right)$

Let $x = 1$, $\implies$, $1 = - B$, $\implies$, $B = - 1$

Let $x = \frac{1}{2}$, $\implies$, $1 = \frac{1}{2} A$, $\implies$, $A = 2$

Therefore,

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

so,

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

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