# Laplace inverse of (e^(−s)(3s^2−s+2))/((s−1)(s^2+1))

Laplace inverse of $\frac{{e}^{-s}\left(3{s}^{2}-s+2\right)}{\left(s-1\right)\left({s}^{2}+1\right)}$
You can still ask an expert for help

## Want to know more about Laplace transform?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Colin Dougherty
I prefer the method of residues to partial fractions. If you recall, the residue of a simple pole ${s}_{0}$ of a function f(s) is
$\underset{s\to {s}_{0}}{lim}\left[\left(s-{s}_{0}\right)f\left(s\right)\right]$
Here
$f\left(s\right)=\frac{3{s}^{2}-s+2}{\left(s-1\right)\left({s}^{2}+1\right)}{e}^{st}$
Therefore are poles at s=1 and $s=±i$. The sum of the residues at these poles are
$2{e}^{t}+\frac{-1-i}{\left(-1+i\right)\left(2i\right)}{e}^{it}+\frac{-1+i}{\left(-1-i\right)\left(-2i\right)}{e}^{-it}=2{e}^{t}+\mathrm{cos}t$
and that is the ILT.