Answer :
[tex]\[ \int_0^t \frac{e^{-t} \sin t}{t} \, dt = \int_0^t e^{-t} \sin t \, dt \]\\\[ \mathcal{L}\{\sin t\} = \frac{1}{s^2 + 1} \]\[ \mathcal{L}\{e^{-t} \sin t\} = \frac{1}{(s+1)^2 + 1} = \frac{1}{s^2 + 2s + 2} \]\[ -\frac{d}{ds} \left( \frac{1}{s^2 + 2s + 2} \right) = \frac{2(s+1)}{(s^2 + 2s + 2)^2} \]\[ \mathcal{L} \left\{ t \int_0^t \frac{e^{-t} \sin t}{t} \, dt \right\} = \frac{2(s+1)}{(s^2 + 2s + 2)^2} \][/tex]
