How do you integrate $int (x-9)/((x+3)(x-7)(x+4))$ using partial fractions?

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Answer 1 · 2016-03-12T05:51:53

$int((x-9)/((x+3)(x-7)(x+4)))dx$

= $132/110ln(x+3)-2/110ln(x-7)-13/11ln(x+4)+c$

To integrate, we should first convert $(x-9)/((x+3)(x-7)(x+4))$ in to partial fractions. Let

$(x-9)/((x+3)(x-7)(x+4))hArrA/(x+3)+B/(x-7)+C/(x+4)$ . Simplifying RHS

= $[A(x-7)(x+4)+B(x+3)(x+4)+C(x+3)(x-7)]/((x+3)(x-7)(x+4))$ or

= $[A(x^2-3x-28)+B(x^2+7x+12)+C(x^2-4x-21)]/((x+3)(x-7)(x+4))$

= $[(A+B+C)x^2+(-3A+7B-4C)x+(-28A+12B-21C)]/((x+3)(x-7)(x+4))$

Hence $A+B+C=0$ , $-3A+7B-4C=1$ and $-28A+12B-21C=-9$

Solving these will give us $A=132/110$ , $B=-2/110$ and $C=-13/11$

Hence $(x-9)/((x+3)(x-7)(x+4))hArr132/(110(x+3))-2/(110(x-7))-13/(11(x+4))$

Hence $int(x-9)/((x+3)(x-7)(x+4))dx$ =

$int[132/(110(x+3))-2/(110(x-7))-13/(11(x+4))]dx$

or

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

Hence, $int[132/(110(x+3))-2/(110(x-7))-13/(11(x+4))]dx$

= $132/110ln(x+3)-2/110ln(x-7)-13/11ln(x+4)+c$

How do you integrate int (x-9)/((x+3)(x-7)(x+4)) int (x-9)/((x+3)(x-7)(x+4)) using partial fractions?