Integration of $\int_0^1 x\log\frac{1+x}{1-x} \, dx$ Wolfram Alpha gives
$$\int_0^1 x\log \frac{1+x}{1-x}\,dx=1$$
I tried integration by part. But there is a convergence issue at $x=1$. Any help? 
 A: First, just calculate the integral on $[0,1-\varepsilon]$:
$$
\lim_{\varepsilon\to 0}\int_{0}^{1-\varepsilon}x\ln\frac{1+x}{1-x}\mathrm{d}x=\boxplus
$$
Since:
$$
\int (1+x)\ln(1+x)=\frac{1}{2} (1+x)^2 \log (1+x)-\frac{(1+x)^2}{4}+C
$$
$$
\int \ln(x+1)=(x+1)\ln (x+1)-(x+1)+C
$$
Using the logarithm-identities:
$$
\boxplus=\lim_{\varepsilon\to 0}\int_{0}^{1-\varepsilon}(x+1)\ln(x+1)-\ln(x+1)+(1-x)\ln(1-x)-\ln(1-x)\mathrm{d}x=
$$
$$
=\lim_{\varepsilon\to 0}\left.\frac{1}{2} x^2 \log \left(\frac{x+1}{1-x}\right)+x+\frac{1}{2} \log (1-x)-\frac{1}{2} \log(x+1)\right|_{0}^{1-\varepsilon}=
$$
$$
=\lim_{\varepsilon\to 0}\left(-\varepsilon +\frac{1}{2} (1-\varepsilon )^2 \log \left(\frac{2-\varepsilon }{\varepsilon
   }\right)-\frac{1}{2} \log (2-\varepsilon )+\frac{\log (\varepsilon )}{2}+1\right).
$$
But $\lim\limits_{\varepsilon\to 0}\varepsilon\log\varepsilon=0$ and expanding gives us:
$$
\lim\limits_{\varepsilon\to 0}\left(\frac{1}{2} \varepsilon ^2 \log (2-\varepsilon )-\frac{1}{2} \varepsilon ^2 \log (\varepsilon )-\varepsilon
   -\varepsilon  \log (2-\varepsilon )+\varepsilon  \log (\varepsilon )+1\right)=1
$$
A: We want to find $\int_0^1 x\log(1+x)\,dx-\int_0^1 x\ln(1-x)\,dx$.
Since the second integral is under suspicion, let us look at it.  Let $u=\ln(1-x)$ and $dv=x\,dx$. Then we have $du=-\frac{1}{1-x}\,dx$, and we can take $v=\frac{x^2}{2}-\frac{1}{2}$. (We used a little trick here!)
So our antiderivative is 
$$\ln(1-x)\left(\frac{x^2}{2}-\frac{1}{2}\right) -\int \frac{1+x}{2}\,dx$$
Note that $\lim_{x\to 1^-}(1-x)\ln(1-x)=0$. It follows that
$$\int_0^1 x\ln(1-x)\,dx=-\int_0^1 \frac{1+x}{2}\,dx=-\frac{3}{4}.$$
The other integral is done the same way, except there is no convergence issue to worry about.
