$\int_0^1 \frac{x^{p}}{x^{p+1}+(1-x)^{p+1}} dx=?$ $$\int_0^1 \frac{x^{p}}{x^{p+1}+(1-x)^{p+1}} dx=?$$
I  tried to use 
$$\int_0^1 \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}} dx=\frac{1}{2}$$ and integration by parts. I  do not know if there is any restriction on p.in original question p=2014,Question from Jalil Hajimir.
 A: Assume $p$ is a positive integer. Of course, you don't need it this strong. 
First, observe that
\begin{align}
I(p) =& \int^1_0 \frac{x^p}{x^{p+1}+(1-x)^{p+1}}\ dx  = \int^1_0 \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ \frac{dx}{x}\\
=&  \int^1_0 \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ \frac{(1-x)}{x} dx+  \int^1_0 \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ dx\ \ \ \ \  (1)\\
=&\ \int^1_0 \frac{(1-x)^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ \frac{x}{1-x} dx+  \int^1_0 \frac{(1-x)^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ dx\ \ \ \ \  (2)
\end{align}
which means
\begin{align}
2I(p) = \int^1_0 f(x)+f(1-x)\ dx +1  = 2\int^1_0 f(x)\ dx +1
\end{align}
where
\begin{align}
f(x) = \frac{x^{p+1}}{x^{p+1}+(1-x)^{p+1}}\ \frac{(1-x)}{x} = \frac{(\frac{1-x}{x})}{1+(\frac{1-x}{x})^{p+1}}.
\end{align}
Hence
\begin{align}
I(p) = \int^\infty_0 \frac{u}{(1+u^{p+1})(1+u)^2} du+\frac{1}{2}.
\end{align}
You can finish the rest using contour integration.
Edit:  I realize I did a lot of unnecessary calculations. In fact, we have
\begin{align}
I(p) =&  \int^1_0 \frac{1}{(1+(\frac{1-x}{x})^{p+1})}\frac{dx}{x} = \int^\infty_0 \frac{du}{(1+u^{p+1})(1+u)}
\end{align}
and
\begin{align}
\lim_{p\rightarrow \infty}I(p)=\lim_{p\rightarrow \infty}\int^\infty_0 \frac{du}{(1+u^{p+1})(1+u)} =\int^1_0 \frac{du}{1+u} = \ln(2).
\end{align}
