baffled with the integral $\int_0^1{e^x \over e^x+1} \, dx$ 
Calculate:
  $$I=\int_0^1{e^x\over e^x+1}dx$$

Personal work:
$$I=\int_0^1{e^x\over e^x+1}dx$$
Let $u=e^x$ hence ${du\over e^x}=dx$
So, $$I=\int_1^e {e^x\over u+1}{du\over e^x}=\int _1^e {1\over u+1}\,du=\cdots$$
I've thought of getting rid of the $u+1$ at the bottom of the fraction, so, $$I=\int_1^e (u+1)^{-1} \, du$$
The problem is this: the anti-derivative of any monomial is given by this type: $$x^n={x^{n+1}\over n+1}+C, C\in \mathbb R.$$ The problem is that $-1+1=0$.
 A: You could apply $u$ substitution by letting $u=e^x+1$, and $du=e^x dx$, this yields, $$\int \frac{du}{u}=\ln(u)=\ln(e^x+1)+C$$
A: The derivative of $\ln(1+x)$ is ${1\over{x+1}}$
A: Let $$u=e^x+1$$
$$I=\int_0^1{e^x\over e^x+1}dx = \int_2^{e+1}{du\over u}=$$
$$ \ln(e+1)-\ln(2) = \ln (\frac {e+1}{2})$$
A: Another approach:
$$
\begin{align}
\int_0^1\frac{e^x}{e^x+1}\,\mathrm{d}x
&=\int_0^1\frac1{1+e^{-x}}\,\mathrm{d}x\tag1\\
&=\int_0^1\sum_{k=0}^\infty(-1)^ke^{-kx}\,\mathrm{d}x\tag2\\
&=\sum_{k=0}^\infty(-1)^k\int_0^1e^{-kx}\,\mathrm{d}x\tag3\\
&=1+\sum_{k=1}^\infty(-1)^k\frac1k\left(1-e^{-k}\right)\tag4\\
&=1-\log(2)+\log\left(1+\frac1e\right)\tag5\\[3pt]
&=\log\left(\frac{e+1}2\right)\tag6
\end{align}
$$
Explanation:
$(1)$: divide numerator and denominator by $e^x$
$(2)$: use the geometric series for $\frac1{1+x}=\sum\limits_{k=0}^\infty x^k$
$(3)\vphantom{\sum\limits_1}$: change order of summation and integration (allowed because the series converges absolutely)
$(4)$: evaluate the integrals (separating the term for $k=0$)
$(5)$: use the series for $\log(1+x)=\sum\limits_{k=1}^\infty(-1)^{k-1}\frac{x^k}k$
$(6)$: combine the terms
A: Hint: the solution is given by $$\int\frac{e^x}{e^x+1}dx=\log(1+e^x)+C$$
A: $$I = \big[\ln (u +1)\big]_{1}^{e} = \ln \frac{e+1}{2}$$
