# How to compute the inverse laplace transform of this term? (-3/(10)s−1/5)/((s+1)^2+1)

How to compute the inverse laplace transform of this term? $\frac{-\frac{3}{10}s-\frac{1}{5}}{\left(s+1{\right)}^{2}+1}$
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indivisast7
$\mathcal{L}\left(y\right)=\frac{-\frac{3}{10}s-\frac{1}{5}}{\left(s+1{\right)}^{2}+1}$
you want to get this into the form. $\frac{a\left(s+1\right)+b}{\left(s+1{\right)}^{2}+1}$
At which point:
$y=a{e}^{-t}\mathrm{cos}t+b{e}^{-t}\mathrm{sin}t$
$\frac{-\frac{3}{10}\left(s+1\right)+\frac{1}{10}}{\left(s+1{\right)}^{2}+1}$
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caritatsjq
Hint. One may write
$\frac{-\frac{3}{10}s-\frac{1}{5}}{\left(s+1{\right)}^{2}+1}=-\frac{3}{10}\frac{\left(s+1\right)}{\left(s+1{\right)}^{2}+1}+\frac{1}{10}\frac{1}{\left(s+1{\right)}^{2}+1}$
then one may apply the properties of the inverse Laplace transform to each term