Answer
$$\mathcal{L}^{-1}\left\{F(s)\right\}=\dfrac{2}{5}e^t-\dfrac{2}{5}e^{-4t} $$
Work Step by Step
Do the Partial Fraction Decomposition to rewrite the function
$$\dfrac{2}{s^2+3s-4}=\dfrac{2}{(s+4)(s-1)}=\dfrac{A}{s-1}+\dfrac{B}{s+4}$$
Multiply both sides and solve for $A$ and $B$.
\begin{align*}
&\Rightarrow \quad2=A(s+4)+B(s-1)\\
\\
s=1&\Rightarrow \quad 2=A(5)+B(0)\Rightarrow \color{red}{A=\dfrac{2}{5}}\\
\\
s=-4&\Rightarrow \quad 2=A(0)+B(-5)\Rightarrow \color{red}{B=-\dfrac{2}{5}}\\
\end{align*}
$$\therefore \dfrac{2}{s^2+3s-4}=\dfrac{{}^{2}{\mskip -5mu/\mskip -3mu}_{5}}{s-1}+\dfrac{-{}^{2}{\mskip -5mu/\mskip -3mu}_{5}}{s+4}$$
Substitute the new function into the inverse Laplace transform and solve using #2 in Table 6.2.1.
\begin{align*}
\mathcal{L}^{-1}\left\{\dfrac{2}{s^2+3s-4}\right\}&=\mathcal{L}^{-1}\left\{\dfrac{2}{(s-1)(s+4)}\right\}\\
\\
&=\mathcal{L}^{-1}\left\{\dfrac{{}^{2}{\mskip -5mu/\mskip -3mu}_{5}}{(s-1)}+\dfrac{-{}^{2}{\mskip -5mu/\mskip -3mu}_{5}}{(s+4)}\right\}\\
\\
&=\dfrac{2}{5}\cdot \mathcal{L}^{-1}\left\{\dfrac{1}{s-1}\right\}-\dfrac{2}{5}\cdot \mathcal{L}^{-1}\left\{\dfrac{1}{s-(-4)}\right\}\\
\\
&=\boxed{\color{blue}{\dfrac{2}{5}e^{t}-\dfrac{2}{5}e^{-4t}}}
\end{align*}
