How do you integrate $\frac{1}{(1 - 2 x) (1 - x)}$ $1 / [ (1-2x)(1-x) ]$ using partial fractions?

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Answer 1 · 2017-01-25T06:12:58

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

Let's start the decomposition into partial fractions

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

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

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

$1 = A (1 - x) + B (1 - 2 x)$ $1=A(1-x)+B(1-2x)$

Let $x = 1$ $x=1$ , $\Rightarrow$ $=>$ , $1 = - B$ $1=-B$ , $\Rightarrow$ $=>$ , $B = - 1$ $B=-1$

Let $x = \frac{1}{2}$ $x=1/2$ , $\Rightarrow$ $=>$ , $1 = \frac{1}{2} A$ $1=1/2A$ , $\Rightarrow$ $=>$ , $A = 2$ $A=2$

Therefore,

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

so,

$\int \frac{d x}{(1 - 2 x) (1 - x)} = 2 \int \frac{d x}{1 - 2 x} - 1 \int \frac{d x}{1 - x}$ $intdx/((1-2x)(1-x))=2intdx/(1-2x)-1intdx/(1-x)$

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

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