Help with an integral with an infinite sum? I have the following, very messy integral:
$$\int_{0}^{1}\sum_{k=1}^{\infty} \frac{(-\frac{e^x}{x+1}+1)^k}{k} dx $$
Neither WolframAlpha nor Mathematica was of any help. I'm not even sure where to start on this integral - I don't believe that it even has a closed form, at least in terms of elementary functions. But I can't prove that. I've been able to manipulate the integrand into a variety of forms, none of which has helped me at all.
Is there a closed form for this integral? If so, how can I find it?
 A: Let $I$ denote the integral given by
$$I=\int_{0}^{1}\sum_{k=1}^{\infty} \frac{(1-\frac{e^x}{x+1})^k}{k} \,dx$$
Note that the series representation for $-\log(1-y)$ is given by
$$-\log(1-y)=\sum_{k=1}^\infty \frac{y^k}{k}$$
for $-1\le y<1$.  Therefore, setting $y=1-\frac{e^x}{x+1}$ we find that 
$$\begin{align}
I&=-\int_0^1 \log\left(\frac{e^x}{x+1}\right)\,dx\\\\
&=\int_0^1 \left(\log(1+x)-x\right)\,dx\\\\
&=2\log(2)-\frac32
\end{align}$$
And we are done!
A: Since $1-\frac {e}{2}<1-\frac {e^x}{x+1}<0$, so you can use this identity
$x+\frac {x^2}{2}+\frac {x^3}{3}........\infty=-ln(1-x)$. So your summation will become $S=-ln(1-(1-\frac {e^x}{x+1}))=-ln(\frac {e^x}{x+1})$. Now you have to integrate $S$ from 0 to 1. So integral would become
 $$I=\int_0^1 (ln(x+1)-x)dx=2ln2-\frac {3}{2}$$
Which is the final answer. 
A: The infinite sum in the integrand is equal to the polylogarithm of $-\frac{e^x}{x+1}+1$ of order 1 (See the infinite sum here). So now we have:
$$\int_{0}^{1}\text{Li}_{1}(-\frac{e^x}{x+1}+1)$$
Which simplifies things a lot. Remember that:
$$\text{Li}_{1}(z)=-\ln(1-z)$$
And this seems to be the direction your teacher/the person who gave you this integral wanted you to take, because $$-\ln(1+\frac{e^x}{x+1}+1)=-\ln(\frac{e^x}{x+1})$$
So now we have manipulated the integral into the following form:
$$\int_{0}^{1}-\ln(\frac{e^x}{x+1})$$
Which is easy enough to solve, because by integrating by parts and simplifying we get:
$$\ln(2)-1+\int_{0}^{1}\frac{x^2}{x+1}$$
