$U_n= \int_0^1 \frac{e^{nx}}{e^x +1} \mathrm d x $ I have this integral in a sequence question 
$$U_n= \int_0^1 \frac{e^{nx}}{e^x +1}  \mathrm dx $$
how to solve it ?
 A: Hint: Because of the use of the term sequence, we assume $n$ is a non-negative integer. Then the following identity will be useful:
$$\frac{e^{nx}}{e^x+1}=e^{(n-1)x}-\frac{e^{(n-1)x}}{e^x+1}.$$
A: Try looking at $$U_{n+1}+U_n = \int_0^1\frac{e^{(n+1)x}+e^{nx}}{e^x+1}\,\mathrm dx$$ and see how that helps simplify the integral.
A: Try this:
Consider the substitution $w=e^x+1$. Then $dw=e^x\,dx=(w-1)\,dx$, and
$$
\int_0^1\frac{e^{nx}}{e^x+1}\,dx=\int_2^{e+1}\frac{(w-1)^n}{w}\cdot\frac{dw}{w-1}=\int_2^{e+1}\frac{(w-1)^{n-1}}{w}\,dw.
$$
Expanding this out with the binomial theorem yields
$$
\int_2^{e+1}\frac{(w-1)^{n-1}}{w}\,dw=\sum_{k=0}^{n-1}\binom{n-1}{k}(-1)^{n-k-1}\int_2^{e+1}\frac{w^k}{w}\,dw
$$
From here, this is manageable.
A: Substitute
$$\begin{gather}t=e^x +1 \;\; \Rightarrow \;\;x=\ln(t-1), \\
x=0 \;\; \Leftrightarrow \;\; t=2,\\
x=1\;\; \Leftrightarrow \;\; t=e+1, \\
dx= \dfrac{dt}{t-1}. \end{gather}$$
Then 
$$\int\limits_{0}^{1}{\dfrac{e^{nx}}{e^x +1} \ dx}=\int\limits_{2}^{e+1}{\dfrac{(t-1)^{n}}{t} \cdot \dfrac{dt}{t-1}}=\int\limits_{2}^{e+1} {\dfrac{(t-1)^{n-1}}{t} \  dt}$$.
A: Or expand in a geometrical series:
$$\int_0^1 dx \frac{e^{n x}}{1+e^x} = \int_0^1 dx \frac{e^{(n-1) x}}{1+e^{-x}} =\sum_{k=0}^{\infty} (-1)^k \int_0^1 dx \, e^{[n-(k-1)] x}$$
The integral is easy, and we get the sum
$$\sum_{k=0}^{\infty} (-1)^k \frac{e^{n-k+1}-1}{n-k+1}$$
That is about all I can do with that.
