Answer
$$\int_{0}^{1} \int_{0}^{e^{v}} \sqrt{1+e^{v}} d w d v= \frac{2}{3}[(1+e) \sqrt{1+e}-2 \sqrt{2}]$$
Work Step by Step
Given $$\int_{0}^{1} \int_{0}^{e^{v}} \sqrt{1+e^{v}} d w d v$$
\begin{aligned}I&=\int_{0}^{1} \int_{0}^{e^{v}} \sqrt{1+e^{v}} d w d v \\
&=\int_{0}^{1}[w \sqrt{1+e^{v}}]_{0}^{e^{v}} d v\\
&=\int_{0}^{1} e^{v} \sqrt{1+e^{v}} d v \end{aligned}
Let
$$1+e^v=z \rightarrow e^v dv =dz$$
at $v=0 \rightarrow z=2$, $v=1 \rightarrow z=1+e$
So, we get
\begin{aligned} I&=\int_{2}^{1+e} \sqrt{z} d z \\ &=\int_{2}^{1+e} z^{1 / 2} d z \\
&=\left[\frac{2}{3} z^{3 / 2}\right]_{2}^{1+e}\\
&= \frac{2}{3}\left[(1+e)^{3 / 2}-2^{3 / 2}\right] \\&= \frac{2}{3}[(1+e) \sqrt{1+e}-2 \sqrt{2}] \end{aligned}
