I decomposed integrand into basic fractions,
(x^2-8x+44)/((x+2)*(x-2)^2)=A/(x+2)+B/(x-2)+C/(x-2)^2x2−8x+44(x+2)⋅(x−2)2=Ax+2+Bx−2+C(x−2)2
(x^2-8x+44)=A*(x-2)^2+B*(x^2-4)+C*(x+2)(x2−8x+44)=A⋅(x−2)2+B⋅(x2−4)+C⋅(x+2)
(x^2-8x+44)=A*(x^2-4x+4)+B*(x^2-4)+C*(x+2)(x2−8x+44)=A⋅(x2−4x+4)+B⋅(x2−4)+C⋅(x+2)
(x^2-8x+44)=(A+B)*x^2+(-4A+C)*x+(4A-4B+2C)(x2−8x+44)=(A+B)⋅x2+(−4A+C)⋅x+(4A−4B+2C)
After equating coefficients, I found A+B=1A+B=1, -4A+C=-8−4A+C=−8 and 4A-4B+2C=444A−4B+2C=44 equations,
After solving system of them simultaneously, I found;
A=4, B=-3A=4,B=−3 and C=8C=8.
Thus,
int(x^2-8x+44)/((x+2)*(x-2)^2)dx∫x2−8x+44(x+2)⋅(x−2)2dx
=int4(dx)/(x+2)-int3(dx)/(x-2)+int8(dx)/(x-2)^2∫4dxx+2−∫3dxx−2+∫8dx(x−2)2
=4Ln(|x+2|)-3Ln(|x-2|)-8/(x-2)+C4ln(|x+2|)−3ln(|x−2|)−8x−2+C
