# Find the inverse Laplace transform of frac{6s+15}{(s^2+25)} s>0

Find the inverse Laplace transform of
$\frac{6s+15}{\left({s}^{2}+25\right)}$
$s>0$
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Step 1
The given function is
$\frac{6s+15}{\left({s}^{2}+25\right)}$
$s>0$
we have to find inverse Laplace Transform of the given function
Step 2
$\frac{\left(6s+15\right)}{\left({s}^{2}+25\right)}=\frac{6s}{\left({s}^{2}+25\right)}+\frac{15}{\left({s}^{2}+25\right)}$
Applying inverse Laplace in both side
${L}^{-1}\left[\frac{\left(6s+15\right)}{\left({s}^{2}+25\right)}\right]=6{L}^{-1}\left[\frac{6s}{\left({s}^{2}+25\right)}\right]+15{L}^{-1}\left[\frac{15}{\left({s}^{2}+25\right)}\right]$
We know
${L}^{-1}\left[\frac{s}{{s}^{2}+{a}^{2}}\right]=\mathrm{cos}\left(at\right)$
${L}^{-1}\left[\frac{1}{{s}^{2}+{a}^{2}}\right]=\frac{1}{a}\mathrm{sin}\left(at\right)$
Then
${L}^{-1}\left[\frac{6s+15}{\left({s}^{2}+25\right)}\right]=6\mathrm{cos}\left(5t\right)+\frac{15}{5}\mathrm{sin}\left(5t\right)$
$⇒{L}^{-1}\left[\frac{6s+15}{\left({s}^{2}+25\right)}\right]=6\mathrm{cos}\left(5t\right)+\frac{15}{5}\mathrm{sin}\left(5t\right)$