Integrate: $\int_0^1\frac{\sqrt{1+e^{-x}}}{e^x}\ dx$ $$\int_0^1\frac{\sqrt{1+e^{-x}}}{e^x}\ dx$$ 
Here is my work. Please let me know if my answer is right or acceptable since maybe I failed to simplify it enough.  
 A: Notice, let $1+e^{-x}=u^2\implies -e^{-x}dx=2udu$
$$\int_{0}^{1}\frac{\sqrt{1-e^{-x}}}{e^x}dx=-\int_{0}^{1}(e^{-x})\sqrt{1-e^{-x}}dx$$
$$=-\int_{\sqrt 2}^{\sqrt{1+e^{-1}}}u(2udu) $$ $$=-2\int_{\sqrt 2}^{\sqrt{1+e^{-1}}} u^2du$$ $$=-2\left[\frac{u^3}{3}\right]_{\sqrt 2}^{\sqrt{1+e^{-1}}}$$ 
$$=\frac{2}{3}\left[2\sqrt 2-\left(1+\frac{1}{e}\right)^{3/2}\right]$$ 
A: You can do this much easier with the substitution $u=1+e^{-x}$. The rest is straightforward.
A: Let $t^2=1+e^{-x}$ for $x \in [0,1]$ Thus, $t \in [1+\frac{1}{e},2]$
$$
\begin{align}
2tdt &=-e^{-x}dx\\
\int_0^1 \frac{\sqrt{1-e^{-x}}}{e^x}dx &=-2\int_2^{1+\frac{1}{e}}t^2dt
\end{align}
$$
A: Hint:
We have
$$
\int_{x} \frac{\sqrt{1+e^{-x}}}{e^{x}} = \int_{x} e^{-x}\sqrt{1+e^{-x}}$$$$ = \int_{x}\frac{-2}{3}D(1 + e^{-x})^{3/2} $$$$= \frac{-2}{3}(1 + e^{-x})^{3/2}
$$
Apply chain rule and fundamental theorem of calculus.
A: Integration is quasi-immediate as $1/e^x=e^{-x}$, which is minus the derivative of the expression under the radical.
Then
$$-\int_0^1\sqrt{1+e^{-x}}\,d(e^{-x})\ dx=-\frac23\left.(1+e^{-x})^{3/2}\right|_0^1=-\frac23((1+e^{-1})^{3/2}-2^{3/2}).$$ 

Even simpler with Zachary Selk's hint:
$$-\int_{u=2}^{1+e^{-1}}\sqrt u\,du.$$
A: Assume that $t=\sqrt{1+e^{-x}}$ then $$dt=\frac{1}{2\sqrt{1+e^{-x}}}(-e^{-x})\ dx$$
 $$\int_0^1\frac{\sqrt{1+e^{-x}}}{e^x}\ dx=\int_{\sqrt 2}^{(1+e^{-1})^{1/2}}\frac{\sqrt{1+e^{-x}}}{e^x}\cdot \frac{2\sqrt{1+e^{-x}}}{(-e^{-x})}\ dt$$ 
$$=-2\int_{\sqrt 2}^{(1+e^{-1})^{1/2}}(1+e^{-x})\ dx$$ 
$$=-2\int_{\sqrt 2}^{(1+e^{-1})^{1/2}}t^2\ dt$$ 
$$=-2\left(\frac{t^3}{3}\right)_{\sqrt 2}^{(1+e^{-1})^{1/2}}$$ 
$$=\frac{4\sqrt 2}{3}-\frac{2}{3}\left(1+\frac{1}{e}\right)^{3/2}$$ 
so it is the answer
A: Rewriting
$$\int_0^1\sqrt{1+e^{-x}}e^{-x}dx=-\int_0^1\sqrt{1+e^{-x}}\,de^{-x}$$ makes the substitution obvious so that you can integrate straight away as
$$\left.-\frac23(1+e^{-x})^{3/2}\right|_0^1=-\frac23((1+e^{-1})^{3/2}-2\sqrt2).$$
