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

Question

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 $-1+1=0$.

Answer 1

The derivative of $\ln(1+x)$ is ${1\over{x+1}}$

Answer 2

Hint: the solution is given by $$\int\frac{e^x}{e^x+1}dx=\log(1+e^x)+C$$

Answer 3

You could apply $u$ substitution by letting $u=e^x+1$, and $du=e^x dx$, this yields, $$\int \frac{du}{u}=\ln(u)=\ln(e^x+1)+C$$

Answer 4

$$I = \big[\ln (u +1)\big]_{1}^{e} = \ln \frac{e+1}{2}$$

Answer 5

Let $$u=e^x+1$$

$$I=\int_0^1{e^x\over e^x+1}dx = \int_2^{e+1}{du\over u}=$$

$$ \ln(e+1)-\ln(2) = \ln (\frac {e+1}{2})$$

Answer 6

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