Question 242318
{{{ (26x^2+208x)/(x^2+1)(x+5)=A/(x+5)+B/(x^2+1)+Cx/(x^2+1)}}}
{{{ 26x^2+208x=A(x^2+1)+B(x+5)+Cx(x+5)}}}
{{{ 26x^2+208x=Ax^2+A+Bx+5B+Cx^2+5Cx}}}
{{{ 26x^2+208x=(A+C)x^2+(B+5C)x+(5B+C)}}}
Comparing terms, 
1){{{ A+C=26}}}
2){{{ A+5B=0}}}
3){{{ B+5C=208}}}
From 1), 
{{{ A=26-C}}}
Substitute into 2),
{{{ 26-C+5B=0}}}
{{{ 5B-C=-26}}}
{{{ 25B-5C=-130}}}
Add this to 3)
{{{ B+5C+25B-5C=-130+208}}}
{{{ 26B=78}}}
{{{ B=3}}}
Then, 
{{{ A+5B=0}}}
{{{ A=-5B=-15}}}
and finally,
{{{ A+C=26}}}
{{{ A=41}}}
{{{ (26x^2+208x)/(x^2+1)(x+5)=-15/(x+5)+3/(x^2+1)+41x/(x^2+1)}}}