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

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Answer 1 · 2016-03-16T13:55:13

#int1/((1+x)(1-2x)]dx#

= #1/3ln(1+x)-1/3ln(1-2x)+c#

To integrate, we should first convert #1/((1+x)(1-2x))# in to partial fractions. Let

#1/((1+x)(1-2x))hArrA/(1+x)+B/(1-2x)#. Simplifying RHS

=#[A(1-2x)+B(1+x)]/((1+x)(1-2x)# or

=#[A-2Ax+B+Bx]/((1+x)(1-2x)#

=#[(B-2A)x+(A+B)]/((x+3)(x-7)(x+4))#

Hence #B-2A=0#, #A+B=1# or #A=1-B#

Hence #B-2(1-B)=0# or #3B=2# or

#B=2/3# and hence #A=1/3#

Hence #1/((1+x)(1-2x))hArr1/(3(1+x))+2/(3(1-2x))#

Hence #int1/((1+x)(1-2x))dx# =

#int[1/(3(1+x))+2/(3(1-2x))]dx#

Now one can use the identity #int(1/(ax+b))dx=1/aln(ax+b)#

Hence, #int[1/(3(1+x))+2/(3(1-2x))]dx#

= #1/3ln(1+x)+2/3xx(-1/2)ln(1-2x)+c#

= #1/3ln(1+x)-1/3ln(1-2x)+c#