$L^2$ norm of the first derivative of Legendre polynomials Recently I have encountered a question while studying the orthogonal properties of Legendre's polynomial

$$
\int_{-1}^{1} P_{n}^{\prime}(x) P_{n}^{\prime}(x)=n(n+1), n\geq1
$$

I have tried the above question with the help of the generating function, but it doesn't help much.
 A: From the recursions relations for the Legendre polynomials here, we have
\begin{equation}
P'_n(x)=2\sum_{j=0}^{\lfloor\frac{n-1}{2}\rfloor} \frac{P_{n-1-2j}(x)}{\|P_{n-1-2j}\|^2}
\end{equation}
for $n\ge1$, where $$\|P_{n}\|^2=\int_{-1}^1\left( P_n(x) \right)^2\,dx=\frac{2}{2n+1}$$
Then the integral can be written as
\begin{align}
I_n&=\int_{-1}^{1} P_{n}^{\prime}(x) P_{n}^{\prime}(x)\,dx\\
&=4\sum_{j=0}^{\lfloor\frac{n-1}{2}\rfloor}\sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor}\int_{-1}^{1}\frac{P_{n-1-2j}(x)}{\|P_{n-1-2j}\|^2}\frac{P_{n-1-2k}(x)}{\|P_{n-1-2k}\|^2}\,dx
\end{align}
Orthogonality of the polynomials imposes $j=k$ to obtain the non-vanishing terms in the summation. The integrals give the norm of remaining polynomials, then
\begin{align}
I_n&=4\sum_{j=0}^{\lfloor\frac{n-1}{2}\rfloor}\int_{-1}^{1}\frac{\left( P_{n-1-2j}(x) \right)^2}{\|P_{n-1-2j}\|^4}\,dx\\
&=4\sum_{j=0}^{\lfloor\frac{n-1}{2}\rfloor}\frac{1}{\|P_{n-1-2j}\|^2}\\
&=2\sum_{j=0}^{\lfloor\frac{n-1}{2}\rfloor}\left( 2n-4j-1 \right)\\
&=n(n+1)
\end{align}
independently of the parity of $n$.
