Answer
$\frac{x+9}{10(x^2+9)}-\frac{1}{10(x+1)}$
Work Step by Step
1. Let $\frac{x}{(x^2+9)(x+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+9}$
2. Combine the right side to get $\frac{Ax^2+9A+Bx^2+Bx+Cx+C}{(x^2+9)(x+1)}=\frac{(A+B)x^2+(B+C)x+9A+C}{(x^2+9)(x+1)}$
3. Compare with step 1, we have $A+B=0, B+C=1, 9A+C=0$,
4. Take the difference of the last two equations to get $9A-B=-1$, add to the 1st equation in step3 to get $10A=-1$, thus $A=-\frac{1}{10}$, $B=\frac{1}{10}$, $C=\frac{9}{10}$,
5. We have $\frac{x}{(x^2+9)(x+1)}=\frac{x+9}{10(x^2+9)}-\frac{1}{10(x+1)}$
