Integrate: $\int_0^1\frac{\sqrt{1+e^{-x}}}{e^x}\ dx$

Question

$$\int_0^1\frac{\sqrt{1+e^{-x}}}{e^x}\ dx$$

Here is my work. Please let me know if my answer is right or acceptable since maybe I failed to simplify it enough.

Answer 1

Notice, let $1+e^{-x}=u^2\implies -e^{-x}dx=2udu$ $$\int_{0}^{1}\frac{\sqrt{1-e^{-x}}}{e^x}dx=-\int_{0}^{1}(e^{-x})\sqrt{1-e^{-x}}dx$$ $$=-\int_{\sqrt 2}^{\sqrt{1+e^{-1}}}u(2udu) $$ $$=-2\int_{\sqrt 2}^{\sqrt{1+e^{-1}}} u^2du$$ $$=-2\left[\frac{u^3}{3}\right]_{\sqrt 2}^{\sqrt{1+e^{-1}}}$$ $$=\frac{2}{3}\left[2\sqrt 2-\left(1+\frac{1}{e}\right)^{3/2}\right]$$

Answer 2

You can do this much easier with the substitution $u=1+e^{-x}$. The rest is straightforward.

Answer 3

Let $t^2=1+e^{-x}$ for $x \in [0,1]$ Thus, $t \in [1+\frac{1}{e},2]$ $$ \begin{align} 2tdt &=-e^{-x}dx\\ \int_0^1 \frac{\sqrt{1-e^{-x}}}{e^x}dx &=-2\int_2^{1+\frac{1}{e}}t^2dt \end{align} $$

Answer 4

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.