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

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

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

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