Question 935798
{{{(x+1)/((x-2)^2(x+4))=A/(x-2)+B/(x-2)^2+C/(x+4)}}}
{{{x+1=A(x-2)(x+4)+B(x+4)+C(x-2)^2}}}
{{{x+1=A(x^2+2x-8)+B(x+4)+C(x^2-4x+4)}}}
{{{x+1=(A+C)x^2+(2A+B-4C)x+(-8A+4B+4C)}}}
So then,
{{{A+C=0}}}
{{{A=-C}}}
.
.
{{{2A+B-4C=1}}}
{{{2(-C)+B-4C=1}}}
1.{{{B-6C=1}}}
.
.
{{{-8A+4B+4C=1}}}
{{{-8(-C)+4B+4C=1}}}
2.{{{4B+12C=1}}}
Multiply eq. 1 by 2 and add to eq. 2,
{{{2B-12C+4B+12C=2+1}}}
{{{6B=3}}}
{{{B=1/2}}}
Then,
{{{1/2-6C=1}}}
{{{-6C=1/2}}}
{{{C=-1/12}}}
{{{A=1/12}}}
So then,
{{{(x+1)/((x-2)^2(x+4))=1/(12(x-2))+1/(2(x-2)^2)-1/(12(x+4))}}}