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

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Answer 1 · 2016-05-14T04:45:02

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

Let us first find partial fractions of #x/((x-1)(x+1))# and for this let

#x/((x-1)(x+1))hArrA/(x-1)+B/(x+1)# or

#x/((x-1)(x+1))hArr(A(x+1)+B(x-1))/((x-1)(x+1))# or

#x/((x-1)(x+1))hArr((A+B)x+(A-B))/((x-1)(x+1))# or

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

Hence #intx/((x-1)(x+1))dx=int[1/(2(x-1))+1/(2(x+1)]dx#

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