# Find the Laplace transform if L\left\{f"(t)\right\}=\arctan\left(\frac{1}{s}\right)

Find the Laplace transform if

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Step 1
Formula used:
$L\left\{f"\left(t\right)\right\}={s}^{2}L\left\{f\left(t\right)\right\}-sf\left(0\right)-{f}^{\prime }\left(0\right)$
Given: $L\left\{{f}^{″}\left(t\right)\right\}=\mathrm{arctan}\left(\frac{1}{s}\right)$
$f\left(0\right)=2,{f}^{\prime }\left(0\right)=1$
To find: $L\left\{f\left(t\right)\right\}$
Step 2
Consider
$L\left\{f"\left(t\right)\right\}={s}^{2}L\left\{f\left(t\right)\right\}-sf\left(0\right)-{f}^{\prime }\left(0\right)$
Substitute all the values we get
$⇒\mathrm{arctan}\left(\frac{1}{s}\right)={s}^{2}L\left\{f\left(t\right)\right\}-2s-1$
$⇒{s}^{2}L\left\{f\left(t\right)\right\}=\mathrm{arctan}\left(\frac{1}{s}\right)+2s+1$
$⇒L\left\{f\left(t\right)\right\}=\frac{\mathrm{arctan}\left(\frac{1}{s}\right)+2s+1}{{s}^{2}}$
$⇒L\left\{f\left(t\right)\right\}=\frac{1}{{s}^{2}}\mathrm{arctan}\left(\frac{1}{s}\right)+\frac{2}{s}+\frac{1}{{s}^{2}}$
Step 3
$⇒L\left\{f\left(t\right)\right\}=\frac{1}{{s}^{2}}\mathrm{arctan}\left(\frac{1}{s}\right)+\frac{2}{s}+\frac{1}{{s}^{2}}$