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,

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

$=-2/(x+1)-ln∣x+1∣+2ln∣x-7∣ + C$

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