Answer
$\frac{2}{3}(1+e)^{\frac{3}{2}}-\frac{4}{3}\sqrt{2}$
Work Step by Step
$\int_{0}^{1}\int_{0}^{e^{v}}\sqrt{1+e^{v}}\, dw\, dv=\int_{0}^{1}\biggl[w\sqrt{1+e^{v}}\biggr]_{0}^{e^v}\, dv=\int_{0}^{1}e^v\sqrt{1+e^v}\, dv$
By letting $$u=1+e^v\,\,\,du=e^v\,dv$$
Then, the equation becomes
$\int_{2}^{1+e}\sqrt{u}\,du=\biggl[\dfrac{2}{3}u^\frac{3}{2}\biggr]_{0}^{1+e}=\frac{2}{3}(1+e)^{\frac{3}{2}}-\frac{4}{3}\sqrt{2}$
