Find laplace transform of the following functions. f(t)=t \sin h(2t)

Step 1
The given function is in the form of ${\displaystyle f\left(t\right)=t^{k}g\left(t\right)}$. The Laplace transform of ${\displaystyle t^{k}g\left(t\right)\text{ is }{(-1)}^k\frac{d^k}{d{s}^k}L\left[g\left(t\right)\right]}$ that is ${\displaystyle L\left[t^{k}g\left(t\right)\right]={(-1)}^k\frac{d^k}{d{s}^k}\left(L\left[g\left(t\right)\right]\right)}$
Compare the function ${\displaystyle f\left(t\right)=t\sin h\left(2t\right)}$ with the ${\displaystyle f\left(t\right)=t^{k}g\left(t\right)}$ we found,
k=1
${\displaystyle g\left(t\right)=\sin h\left(2t\right)}$
Since, ${\displaystyle L\left[t^{k}g\left(t\right)\right]={(-1)}^k\frac{d^k}{d{s}^k}\left(L\left[g\left(t\right)\right]\right)}$ therefore ${\displaystyle L\left[t\sin h2t\right]={(-1)}^1\frac{d}{ds}\left(L\left[\sin h2t\right]\right)\dot{s}\left(1\right)}$
Step 2
Since ${\displaystyle L\left[\sin h\left(at\right)\right]=\frac{a}{s^2-a^2}}$ , So use this find ${\displaystyle L\left[\sin h\left(2t\right)\right]}$
${\displaystyle L\left[\sin h\left(2t\right)\right]=\frac{2}{s^2-2^2}}$
${\displaystyle =\frac{2}{s^2-4}\dot{s}\left(2\right)}$
From (1) and (2) , ${\displaystyle L\left[t\sin \left(2t\right)\right]=-\frac{d}{ds}\left[\frac{2}{s^2-4}\right]}$.

Calculate ${\displaystyle \frac{d}{ds}\left[\frac{2}{s^2-4}\right]}$
${\displaystyle \frac{d}{ds}\left[\frac{2}{s^2-4}\right]=\frac{-2}{{(s^2-4)}^2}\cdot 2s}$
${\displaystyle =-\frac{4s}{{(s^2-4)}^2}}$
Therefore

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