Question

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

Answer 1

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`

Explanation:

I decomposed integrand into basic fractions.