How do i find the inverse laplace? $$
F(s) = \frac{2s-1}{s^2(s+1)^3}
$$
If I try to use partial fractions, I end up with 8 constants to solve for!
Is there some shortcut I'm not seeing? Am I supposed to simplify it first? Am I even doing the partial decomposition right?
 A: You should have five.
$$\displaystyle F(s) = \frac{2s-1}{s^2(s+1)^3} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+1} + \frac{D}{(s+1)^2} + \frac{E}{(s+1)^3}$$
This will produce:


*

*$A = 5$

*$B = -1$

*$C = -5$

*$D = -4$

*$E = -3$


The inverse Laplace transform, using this table is:
$$-\dfrac{1}{2} e^{-t} (3~ t^2+2~ t~ e^t+ 8~t-10~ e^t+10)$$
A: Alternatively, if you know some complex analysis, the ILT is the sum of the residues of $F(s) e^{s t}$ at the poles of $F$.  There is a double pole at $s=0$ and a triple pole at $s=-1$, so that 
$$f(t) = \left [ \frac{d}{ds} \frac{(2 s-1) e^{s t}}{(s+1)^3}  \right ]_{s=0} + \left [ \frac12 \frac{d^2}{ds^2} \frac{(2 s-1) e^{s t}}{s^2}  \right ]_{s=-1}$$
I leave the details to the reader; the result is
$$-\frac{1}{2} e^{-t} \left(3 t^2+8 t+10\right)-t+5$$
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{{\rm F}\pars{s} = {2s - 1 \over s^{2}\pars{s + 1}^{3}}.\quad
{\cal F}\pars{t}
=
\int_{\gamma - \ic\infty}^{\gamma + \ic\infty}{\rm F}\pars{s}\expo{st}\,
{\dd s \over 2\pi\ic}\quad\mbox{where}\quad \Re\gamma > 0}$

\begin{align}
{\cal F}\pars{t}
&=
\int_{\gamma - \ic\infty}^{\gamma + \ic\infty}\,{2s - 1 \over s^{2}
\pars{s + 1}^{3}}\,\expo{st}\,{\dd s \over 2\pi\ic}
\\[3mm]&=
\lim_{s \to 0}\partiald{}{s}\bracks{{\pars{2s - 1}\expo{st} \over \pars{s + 1}^{3}}}
+
{1 \over 2}\,\lim_{s \to\ -1}\partiald[2]{}{s}
\bracks{{\pars{2s - 1}\expo{st} \over s^{2}}}\tag{1}
\end{align}

\begin{align}
&\lim_{s \to 0}\partiald{}{s}\bracks{{\pars{2s - 1}\expo{st} \over \pars{s + 1}^{3}}}
=
\lim_{s \to 0}\bracks{%
{2\expo{st} + \pars{2s - 1}\expo{st}t \over \pars{s + 1}^{3}}
-
3\,{\pars{2s - 1}\expo{st} \over \pars{s + 1}^{4}}}
\\[3mm]&= \pars{2 - t} + 3 = 5 - t\tag{2}
\\[3mm]&
{1 \over 2}\,\lim_{s \to\ -1}\partiald[2]{}{s}
\bracks{{\pars{2s - 1}\expo{st} \over s^{2}}}
=
{1 \over 2}\,\lim_{s \to 0}\partiald[2]{}{s}
\braces{{\bracks{2\pars{s - 1} - 1}\expo{\pars{s - 1}t} \over \pars{s - 1}^{2}}}
\\[3mm]&=
{1 \over 2}\,\expo{-t}\lim_{s \to 0}\partiald[2]{}{s}
\bracks{{\pars{2s - 3}\expo{st} \over \pars{s - 1}^{2}}}
=
{1 \over 2}\,\expo{-t}\underbrace{\pars{-3t^{2} - 8t - 10}}_{\mbox{From WA}}\tag{3}
\end{align}
By replacing $\pars{2}$ and $\pars{3}$ in $\pars{1}$, we get:
$$\color{#0000ff}{\large%
\int_{\gamma - \ic\infty}^{\gamma + \ic\infty}\,{2s - 1 \over s^{2}
\pars{s + 1}^{3}}\,\expo{st}\,{\dd s \over 2\pi\ic}
=
-\,{3 \over 2}\expo{-t}\pars{t^{2} + {8 \over 3}\,t + {10 \over 3}} - t + 5}\,,
\quad \Re\gamma > 0
$$
