Codex @codex  0

Differentiation under the integral gives $I\prime \prime ={\int }_0^1{y}^4{e}^{xy}dy$. Integrating the derivative of $e^{xy}(x{y}^3-3y^2)$ and then once more yields $x^2{\int }_0^1{y}^4{e}^{xy}dy-12{\int }_0^1{y}^2{e}^{xy}dy=e^x(x-4)$.