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

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

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Answer 1 · 2016-03-09T12:37:10

$\frac{1}{(1 + x) (1 - 2 x)} \Leftrightarrow \frac{1}{3 (1 + x)} + \frac{2}{3 (1 - 2 x)}$ $1/((1+x)(1-2x))hArr1/(3(1+x))+2/(3(1-2x))$

Explanation:

Let $\frac{1}{(1 + x) (1 - 2 x)} \Leftrightarrow \frac{A}{1 + x} + \frac{B}{1 - 2 x}$ $1/((1+x)(1-2x))hArrA/(1+x)+B/(1-2x)$

Simplifying RHS, it is equal to

$\frac{1}{(1 + x) (1 - 2 x)} \Leftrightarrow \frac{A (1 - 2 x) + B (1 + x)}{(1 + x) (1 - 2 x)}$ $1/((1+x)(1-2x))hArr(A(1-2x)+B(1+x))/((1+x)(1-2x))$ or

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

i.e. $B - 2 A = 0$ $B-2A=0$ and $A + B = 1$ $A+B=1$

From first we get $B = 2 A$ $B=2A$ and putting this in second we get

$A + 2 A = 1$ $A+2A=1$ or $3 A = 1$ $3A=1$ or $A = \frac{1}{3}$ $A=1/3$ . As $B = 2 A = 2 \times \frac{1}{3} = \frac{2}{3}$ $B=2A=2xx1/3=2/3$ , we have

$\frac{1}{(1 + x) (1 - 2 x)} \Leftrightarrow \frac{1}{3 (1 + x)} + \frac{2}{3 (1 - 2 x)}$ $1/((1+x)(1-2x))hArr1/(3(1+x))+2/(3(1-2x))$

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

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How do you express 1/[(1+x)(1-2x)]1(1+x)(1−2x)1/[(1+x)(1-2x)] in partial fractions?

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