Answer
$\frac{3x+1}{(x-5)(x+3)}=\frac{2}{x-5}+\frac{1}{x+3}$
Work Step by Step
$\frac{3x+1}{x^2-2x-15}$, factorising $x^2-2x-15=(x-5)(x+3)$,
thus,
$\frac{3x+1}{(x-5)(x+3)}=\frac{A}{x-5}+\frac{B}{x+3}$,
$A(x+3)+B(x-5)$,
$Ax+3A+Bx-5B$,
$(A+B)x+(3A-5B)=3x+1$,
$\begin{cases}
A+B=3\\
3A-5B=1
\end{cases}$,
Multiplying Equation 1 by 5 and adding it to Equation 2.
$\begin{cases}
5A+5B=15\\
3A-5B=1\\
-- -- -- --\\
8A+0B=16
\end{cases}$
thus, $A=2$, substituting back into the Equation, $2+B=3, B=1$.
Therefore,
$\frac{3x+1}{(x-5)(x+3)}=\frac{2}{x-5}+\frac{1}{x+3}$
