Answer
$\frac{2}{x+3}+\frac{7}{x-1}$
Work Step by Step
1. Factor the denominator as $x^2+2x-3=(x+3)(x-1)$
2. Assume $\frac{9x+19}{x^2+2x-3}=\frac{A}{x+3}+\frac{B}{x-1}$
3. Evaluate the right side, we have $RHS=\frac{Ax-A+Bx+3B}{x^2+2x-3}=\frac{(A+B)x-A+3B}{x^2+2x-3}$
4. Compare the above with the left hand side to get $\begin{cases} A+B=9 \\ -A+3B=19 \end{cases}$
5. Add the two equations to get $4B=28$, thus $B=7$ and $A=9-B=2$
6. The partial fraction decomposition is $\frac{9x+19}{x^2+2x-3}=\frac{2}{x+3}+\frac{7}{x-1}$
