Answer to Question #233092 in Math for abi

Question

L {t² cos t}

Answer 1

$L[t^2cos t]=(-1)^2\frac{d^2}{ds^2}L[cost]=\frac{d^2}{ds^2}(\frac{s}{s^2+1})=\frac{2s(s^2-3)}{(s^2+1)^3}.$