inverse laplacian transform of $\frac{4s + 6}{(s+1)^2}$

Question

I'm having trouble figuring out how to solve this inverse laplacian transform-

$$\frac{4s + 6}{(s+1)^2}$$

Can anyone give me a few pointers? Thanks in advance.

Answer 1

Just to add to @jack-daurizio's answer, this arises from the following:

$$\begin{align} \frac{4s+6}{(s+1)^{2}} &\,=\, \frac{4(s+1)+2}{(s+1)^{2}} \,=\, \frac{4(s+1)}{(s+1)^{2}} + \frac{2}{(s+1)^{2}} \,=\, \frac{4}{s+1} + \frac{2}{(s+1)^{2}} \end{align}.$$

Then proceed as mentioned!

Answer 2

Pretty straightforward: the inverse Laplace transform of $\frac{1}{s+1}$ is $e^{-t}$, hence the inverse Laplace transform of $\frac{1}{(s+1)^2}$ is $t e^{-t}$ and

$$ \mathcal{L}^{-1}\left(\frac{4s+6}{(s+1)^2}\right) = 4\mathcal{L}^{-1}\left(\frac{1}{s+1}\right)+2\mathcal{L}^{-1}\left(\frac{1}{(s+1)^2}\right) = \color{red}{(4+2t)e^{-t}} $$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{\pars{-1}^{+} - \infty\ic}^{\pars{-1}^{+} + \infty\ic} {4s + 6 \over \pars{s + 1}^{2}}\,\expo{ts}\,{\dd s \over 2\pi\ic} & = \bracks{t > 0}\braces{2\pi\ic\lim_{s\ \to\ -1} \partiald{}{s}\bracks{\pars{s + 1}^{2}\, {4s + 6 \over \pars{s + 1}^{2}}\,\expo{ts}\,{1 \over 2\pi\ic}}} \\[5mm] & = \bracks{t > 0}\lim_{s\ \to\ -1} \partiald{}{s}\bracks{\pars{4s + 6}\,\expo{ts}} \\[5mm] & = \bracks{t > 0}\lim_{s\ \to\ -1} \bracks{4\expo{ts} + \pars{4s + 6}\,\expo{ts}t} \\[5mm] & = \bbx{\bracks{t > 0}\bracks{2\pars{t + 2}\expo{-t}}} \end{align}