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))#