Answer
$$\int_{0}^{\ln 2} \int_{0}^{1} x y e^{y^{2} x} \ d y \ d x =\frac{1-\ln2}{2} $$
Work Step by Step
Given $$\int_{0}^{\ln 2} \int_{0}^{1} x y e^{y^{2} x} d y \ d x$$
So, we have
\begin{aligned}
I&=\int_{0}^{\ln 2} \int_{0}^{1} x y e^{y^{2} x} \ \ d y\ d x\\
&=\frac{1}{2}\int_{0}^{\ln 2} \int_{0}^{1} 2x y \ e^{y^{2} x} d y \ d x\\
&=\frac{1}{2}\int_{0}^{\ln 2} \ e^{y^{2} x}|_{0}^{1} d x\\
&=\frac{1}{2}\int_{0}^{\ln 2} \left( \ e^{ x}-1\right) d x\\
&=\frac{1}{2} \left( \ e^{ x}-x\right)_{0}^{\ln 2}\\
&=\frac{1}{2} \left( \ e^{ \ln 2}-\ln2\right)-\frac{1}{2} \left(1-0\right)\\
&=\frac{1}{2} \left(2-\ln2-1\right) \\
&=\frac{1-\ln2}{2} \\
\end{aligned}
