Solution
= Solution
<Differentiation> under the <integral> gives $I\prime\prime=\int_0^1y^4e^{xy}dy$. Integrating the <derivative> of $e^{xy}(xy^3-3y^2)$ and then once more yields $x^2\int_0^1y^4e^{xy}dy-12\int_0^1y^2e^{xy}dy=e^x(x-4)$.
= Solution
<Differentiation> under the <integral> gives $I\prime\prime=\int_0^1y^4e^{xy}dy$. Integrating the <derivative> of $e^{xy}(xy^3-3y^2)$ and then once more yields $x^2\int_0^1y^4e^{xy}dy-12\int_0^1y^2e^{xy}dy=e^x(x-4)$.