# Find Laplace transform for frac{s+8}{s^2+4s+5}

Find Laplace transform for
$\frac{s+8}{{s}^{2}+4s+5}$
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Step 1
Formula Used:
$L\left\{{e}^{at}\mathrm{sin}bt\right\}=\frac{b}{\left(s-a{\right)}^{2}+{b}^{2}}$
$L\left\{{e}^{at}\mathrm{cos}bt\right\}=\frac{s-a}{\left(s-a{\right)}^{2}+{b}^{2}}$
Step 2

$=\frac{s+8}{\left(s+2{\right)}^{2}+1}$
$=\frac{s+2+6}{\left(s+2{\right)}^{2}+1}$
$=\frac{\left(s+2\right)}{\left(s+2{\right)}^{2}+1}+\frac{6}{\left(s+2{\right)}^{2}+1}$
Step 3

We now apply Inverse Laplace Transform, we get
${L}^{-1}\left\{F\left(s\right)\right\}={L}^{-1}\left\{\frac{\left(s+2\right)}{\left(s+2{\right)}^{2}+1}\right\}+6{L}^{-1}\left\{\frac{1}{\left(s+2{\right)}^{2}+1}\right\}$

$={e}^{-2t}\left(\mathrm{cos}t+6\mathrm{sin}t\right)$
Step 4
: The inverse Laplace Transform of F(s) is
$f\left(t\right)={e}^{-2t}\left(\mathrm{cos}t+6\mathrm{sin}t\right)$