$\mathcal{L}^{-1} \left[ \dfrac{s - 5}{s^2 + s - 6} \right] = \mathcal{L}^{-1} \left[ \dfrac{s - 5}{(s + 3)(s - 2)} \right]$
For
$\dfrac{s - 5}{(s + 3)(s - 2)} = \dfrac{A}{s + 3} + \dfrac{B}{s - 2}$
$s - 5 = A(s - 2) + B(s + 3)$
set $s = -3;$ $-8 = -5A;$ $A = 8/5$
set $s = 2;$ $-3 = 5B;$ $B = -3/5$
Thus,
$\mathcal{L}^{-1} \left[ \dfrac{s - 5}{s^2 + s - 6} \right] = \mathcal{L}^{-1} \left[ \dfrac{8/5}{s + 3} + \dfrac{-3/5}{s - 2} \right]$
$\mathcal{L}^{-1} \left[ \dfrac{s - 5}{s^2 + s - 6} \right] = \dfrac{8}{5}\mathcal{L}^{-1} \left( \dfrac{1}{s + 3} \right) - \dfrac{3}{5} \mathcal{L}^{-1} \left( \dfrac{1}{s - 2} \right)$
$\mathcal{L}^{-1} \left[ \dfrac{s - 5}{s^2 + s - 6} \right] = \frac{8}{5}e^{-3t} - \frac{3}{5} e^{2t}$ answer
