Legendre functions $Q_n(x)$ of the second kind Legendre functions $Q_n(x)$ of the second kind
\begin{equation*}
Q_n(x)=P_n(x) \int \frac{1}{(1-x^2)\cdot P_n^2(x)}\, \mathrm{d}x
\end{equation*}
what to do after this step?
how can I complete ?
I need to reach this formula
\begin{equation*}
Q_n(x)=\frac{1}{2} P_n(x)\ln\left( \frac{ 1+x}{1-x}\right)
\end{equation*}
 A: So you want to prove that 
$$Q_n(x)= P_n(x) \int^x \frac{1}{(1-x^2) \, \left[P_n(x)\right]^2} \, dx = \frac{1}{2} P_n(x) \ln\left(\frac{1+x}{1-x}\right)$$.
Let's first compute for the case $n=0$, knowing that $P_0(x) = 1$ we would have
$$Q_0(x) = \int^x \frac{1}{(1-x^2)} \, dx = \frac{1}{2} \int \left[\frac{1}{1+x} + \frac{1}{1-x} \right] = \frac{1}{2} \, \ln \left(\frac{1+x}{1-x} \right)$$
where we decomposed the fraction to compute the integral.
Let us now compute for $n=1$, where $P_1(x) = x$.
$$Q_1(x) = x \int^x \frac{1}{(1-x^2)\,x^2} \, dx = \frac{x}{2} \, \ln \left(\frac{1+x}{1-x} \right) - 1$$
where we decomposed the fraction again to compute for the integral.
For $n=2$ we use the same process again, leading us to the result
$$Q_2(x) = \frac{1}{2} P_2(x) \ln \left( \frac{1+x}{1-x} \right) - \frac{3x}{2}$$
So it seems that it only works for $n=0$ and that the true formula is 
$$Q_n(x) = \frac{1}{2} P_n(x) \ln \left( \frac{1+x}{1-x} \right) - C_n(x)$$
for some value of $C_n(x)$.

As an alternative you could just use the recurrence formulas for $Q_n(x)$ which are
$$(n+1)Q_{n+1}(x) - (2n+1)x \, Q_n(x) + n \, Q_{n-1}(x) = 0$$
$$(2n+1)Q_n(x) = Q'_{n+1}(x) -Q'_{n-1}(x)$$
To show that
$$Q_n(x) = \frac{1}{2} P_n(x) \ln \left(\frac{1+x}{1-x}\right) - \frac{2n-1}{n}P_{n-1}(x) - \frac{2n-5}{3(n-1)} P_{n-3}(x) - \cdots$$

If you want to read more about the subject then I suggest you take a look at some mathematical physics books like Mathematical Methods for Physicists by Arfken (Some of my friends say that they dumbed down the seventh edition, but I still use the seventh anyways) or Mathematical Methods in the Physical Sciences by Boas.
