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

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Answer 1 · 2016-03-09T12:37:10

$1/((1+x)(1-2x))hArr1/(3(1+x))+2/(3(1-2x))$

Let $1/((1+x)(1-2x))hArrA/(1+x)+B/(1-2x)$

Simplifying RHS, it is equal to

$1/((1+x)(1-2x))hArr(A(1-2x)+B(1+x))/((1+x)(1-2x))$ or

$1/((1+x)(1-2x))hArr((B-2A)x+(A+B))/((1+x)(1-2x))$

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

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

$A+2A=1$ or $3A=1$ or $A=1/3$ . As $B=2A=2xx1/3=2/3$ , we have

$1/((1+x)(1-2x))hArr1/(3(1+x))+2/(3(1-2x))$

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