Question

inverse laplace transform of 3s+5/(s^2-6s+25)

Answer 1

Complete the square in the denominator first:
\[\begin{align*}s^2-6s+25&=s^2-6s+9+16\\
&=(s-3)^2+16
\end{align*}\]
\[\begin{align*}F(s)&=\frac{3s+5}{(s-3)^2+16}\\
&=\frac{3s-9+14}{(s-3)^2+16}\\
&=\frac{3(s-3)}{(s-3)^2+16}+\frac{14}{(s-3)^2+16}\\
\mathcal{L}^{-1}\left\{F(s)\right\}&=3\mathcal{L}^{-1}\left\{\frac{s-3}{(s-3)^2+16}\right\}+\frac{14}{4}\mathcal{L}^{-1}\left\{\frac{4}{(s-3)^2+16}\right\}\\
f(t)&=3e^{3t}\cos4t+\frac{14}{4}e^{3t}\sin4t
\end{align*}\]

Answer 2

A "missing" step in red, added to clarify where the \(e^{3t}\) comes from.
\[\begin{align*}F(s)&=\frac{3s+5}{(s-3)^2+16}\\
&=\frac{3s-9+14}{(s-3)^2+16}\\
&=\frac{3(s-3)}{(s-3)^2+16}+\frac{14}{(s-3)^2+16}\\
\mathcal{L}^{-1}\left\{F(s)\right\}&=3\mathcal{L}^{-1}\left\{\frac{s-3}{(s-3)^2+16}\right\}+\frac{14}{4}\mathcal{L}^{-1}\left\{\frac{4}{(s-3)^2+16}\right\}\\
&=\color{red}{3e^{3t}\mathcal{L}^{-1}\left\{\frac{s}{s^2+16}\right\}+\frac{14}{4}e^{3t}\mathcal{L}^{-1}\left\{\frac{4}{s^2+16}\right\}}\\
f(t)&=3e^{3t}\cos4t+\frac{14}{4}e^{3t}\sin4t
\end{align*}\]