How do you integrate $int (x^2 - 8x + 44) / ((x + 2) (x - 2)^2)$ using partial fractions?

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Answer 1 · 2017-07-11T12:08:36

I decomposed integrand into basic fractions,

$(x^2-8x+44)/((x+2)*(x-2)^2)=A/(x+2)+B/(x-2)+C/(x-2)^2$

$(x^2-8x+44)=A*(x-2)^2+B*(x^2-4)+C*(x+2)$

$(x^2-8x+44)=A*(x^2-4x+4)+B*(x^2-4)+C*(x+2)$

$(x^2-8x+44)=(A+B)*x^2+(-4A+C)*x+(4A-4B+2C)$

After equating coefficients, I found $A+B=1$ , $-4A+C=-8$ and $4A-4B+2C=44$ equations,

After solving system of them simultaneously, I found;

$A=4, B=-3$ and $C=8$ .

Thus,

$int(x^2-8x+44)/((x+2)*(x-2)^2)dx$

= $int4(dx)/(x+2)-int3(dx)/(x-2)+int8(dx)/(x-2)^2$

= $4Ln(|x+2|)-3Ln(|x-2|)-8/(x-2)+C$

I decomposed integrand into basic fractions.

How do you integrate int (x^2 - 8x + 44) / ((x + 2) (x - 2)^2)int (x^2 - 8x + 44) / ((x + 2) (x - 2)^2) using partial fractions?