Laplace

$\begin{eqnarray}
\mathcal{L} (\sin t) &=& \dfrac{1}{s^2+1}\\
\mathcal{L} (t \sin t) &=& -\left(\dfrac{1}{s^2+1} \right)' = \dfrac{2s}{(s^2+1)^2}\\
\mathcal{L}(te^{5t} \sin t) &=& \dfrac{2(s - 5)}{\left[(s-5)^2 + 1 \right]^2}\\
\mathcal{L}(te^{5t+1} \sin t) &=& \boxed{\dfrac{2e(s-5)}{\left[(s-5)^2+1 \right]^2}}
\end{eqnarray}$