Answer
$\frac{4\sqrt{2}}{3}-\frac{2(1+\frac{1}{e})^\frac{3}{2}}{3}$
Work Step by Step
Let $u=1+e^{-x}$. Then $du=-e^{-x}=-\frac{1}{e^x}$, and $-du=\frac{1}{e^x}$.
$\int_0^1\frac{\sqrt{1+e^{-x}}}{e^x}dx$
$=\int_{1+e^{-0}}^{1+e^{-1}}\sqrt{u}*(-1)du$
$=-\int_2^{1+\frac{1}{e}}u^{\frac{1}{2}}du$
$=-\left.\left(\frac{u^\frac{3}{2}}{\frac{3}{2}}\right)\right|_2^{1+\frac{1}{e}}$
$=-\left.\left(\frac{2u^\frac{3}{2}}{3}\right)\right|_2^{1+\frac{1}{e}}$
$=-\left(\frac{2(1+\frac{1}{e})^\frac{3}{2}}{3}-\frac{2*2^\frac{3}{2}}{3}\right)$
$=\frac{2*2^\frac{3}{2}}{3}-\frac{2(1+\frac{1}{e})^\frac{3}{2}}{3}$
$=\frac{2*2\sqrt{2}}{3}-\frac{2(1+\frac{1}{e})^\frac{3}{2}}{3}$
$=\frac{4\sqrt{2}}{3}-\frac{2(1+\frac{1}{e})^\frac{3}{2}}{3}$