$f(t) = t \, \sin t$ .......... $f(0) = 0$
$f'(t) = t \, \cos t + \sin t$ .......... $f'(0) = 0$
$f''(t) = (-t \, \sin t + \cos t) + \cos t = -t \, \sin t + 2\cos t$
$\mathcal{L} \left\{ f''(t) \right\} = s^2 \mathcal{L} \left\{ f(t) \right\} - s \, f(0) - f'(0)$
$\mathcal{L} (-t \, \sin t + 2\cos t) = s^2 \mathcal{L} (t \, \sin t) - s \, (0) - (0)$
$-\mathcal{L} (t \, \sin t) + 2\mathcal{L} (\cos t) = s^2 \mathcal{L} (t \, \sin t)$
$2\mathcal{L} (\cos t) = s^2 \mathcal{L} (t \, \sin t) + \mathcal{L} (t \, \sin t)$
$s^2 \mathcal{L} (t \, \sin t) + \mathcal{L} (t \, \sin t) = 2\mathcal{L} (\cos t)$
$(s^2 + 1) \mathcal{L} (t \, \sin t) = 2\mathcal{L} (\cos t)$
$(s^2 + 1) \mathcal{L} (t \, \sin t) = 2 \left( \dfrac{s}{s^2 + 1^2} \right)$
$(s^2 + 1) \mathcal{L} (t \, \sin t) = \dfrac{2s}{s^2 + 1}$
$\mathcal{L} (t \, \sin t) = \dfrac{2s}{(s^2 + 1)(s^2 + 1)}$
$\mathcal{L} (t \, \sin t) = \dfrac{2s}{(s^2 + 1)^2}$ answer
