How do you integrate #(x^2+12x-5)/((x+1)^2(x-7))# using partial fractions?

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Answer 1 · 2016-11-27T07:53:05

The answer is #=-2/(x+1)-ln∣x+1∣+2ln∣x-7∣ + C#

Let's do the decomposition into partial fractions

#(x^2+12x-5)/((x+1)^2(x-7))=A/(x+1)^2+B/(x+1)+C/(x-7)#

#=(A(x-7)+B(x+1)(x-7)+C(x+1)^2)/((x+1)^2(x-7))#

Therefore,

#x^2+12x-5=A(x-7)+B(x+1)(x-7)+C(x+1)^2#

Let #x=-1#, #=>#, #-16=-8A#

Coefficients of #x^2#, #1=B+C#

Let #x=7#, #=>#, #128=64C#

#C=2#

#B=1-C=1-2=-1#

#A=2#

#(x^2+12x-5)/((x+1)^2(x-7))=2/(x+1)^2-1/(x+1)+2/(x-7)#

So,

#int((x^2+12x-5)dx)/((x+1)^2(x-7))=int(2dx)/(x+1)^2-intdx/(x+1)+int(2dx)/(x-7)#

#=-2/(x+1)-ln∣x+1∣+2ln∣x-7∣ + C#