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

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

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