How to do this Orthogonality Integral? I am solving a question, where
$$ p_1(x) = x , Q_0(x) = \frac {\ln \frac {1+x} {1-x}}{2}$$
are solutions of legedre's differential equation corresponding to different eigen values.
I have to evaluate their orthogonality integral
$$ \int \limits_{-1}^{1} x \frac {\ln \frac {1+x} {1-x}}{2}$$
While doing this, I came to a point where I have to do
$$ \int \limits_{-1}^{1} \ln(1+x)dx $$
which leads to 
$$ [(1+x)\ln(1+x) - (1+x)]_{-1}^{1} $$ which gives a term
$$ (1-1)\ln(1-1) -(1-1) $$ 
What is it's value ? Is it zero? The space here is Hilbert space.

Ignoring limits,
$$
    \int x\ln\left(\frac{1+x}{1-x}\right) dx \\
   = \frac{1}{2}\int \frac{d}{dx}(x^2-1)\ln\left(\frac{1+x}{1-x}\right)dx \\
   = \frac{1}{2}(x^2-1)\ln\left(\frac{1+x}{1-x}\right)-\frac{1}{2}\int(x^2-1)\frac{d}{dx}\ln\left(\frac{1+x}{1-x}\right)dx \\
   = \frac{1}{2}(x^2-1)\ln\left(\frac{1+x}{1-x}\right)-\frac{1}{2}\int(x^2-1)\left[\frac{1}{1+x}+\frac{1}{1-x}\right]dx \\
   = \frac{1}{2}(x^2-1)\ln\left(\frac{1+x}{1-x}\right)-\int(x^2-1)\frac{1}{1-x^2}dx \\
  = \frac{1}{2}(x^2-1)\ln\left(\frac{1+x}{1-x}\right)+x+C.
$$
Evaluating the right side between $1^{-}$ and $(-1)^{+}$ gives 2. Therefore,
$$
    \langle p_1,Q_0\rangle = \lim_{\epsilon\downarrow 0}\int_{-1+\epsilon}^{1-\epsilon}p_1(x)Q_0(x)dx = 1.
$$
