Answer
$\frac{3}{x+4}+\frac{7}{x-5}$
Work Step by Step
1. Factor the denominator to get $x^2-x-20=(x-5)(x+4)$
2. Assume $\frac{10x+13}{x^2-x-20}=\frac{A}{x+4}+\frac{B}{x-5}$
3. We have $RHS=\frac{Ax-5A+Bx+4B}{(x-5)(x+4)}=\frac{(A+B)x-5A+4B}{(x-5)(x+4)}$
4. Compare steps 2 and 3 to get $\begin{cases} A+B=10 \\ -5A+4B=13 \end{cases}$
5. Multiply 5 to the first equation and add the result to the second to get $9B=63$, thus $B=7$ and $A=3$
6. We have $\frac{10x+13}{x^2-x-20}=\frac{3}{x+4}+\frac{7}{x-5}$
