I was trying to evaluate -
$\int_{-1}^{1} x P_{n}(x) P_{m}(x) dx$
Point is there is a $x$ in the integrand, otherwise, it is a well known integral which results to $\frac{2}{2n+1}$ if $m=n$ and $0$ otherwise that is the Legendre Polynomials are orthogonal.
EDIT It is given that the integral results to $\frac{2(n+1)}{(2n+1)(2n+3)} \delta_{m,n+1} + \frac{2n}{(2n+1)(2n-1)} \delta_{m,n-1}$
But how to solve this integral?
A nice idea here is to use a nice recursion for Legendre Polynomials, namely $$(1-x^2)P_n'(x) = (n+1)\left(xP_n(x) - P_{n+1}(x)\right)$$ We can use this by first applying integration by parts to get $$I=\int_{-1}^1 xP_n(x)P_m(x)dx = \frac{1-x^2}{2}P_n(x)P_m(x)\bigg|_{-1}^1+\frac{1}{2}\int_{-1}^1(1-x^2)(P_n'(x)P_m(x) + P_m'(x)P_n(x))dx$$ Since $1-x^2$ is $0$ for $x=1,-1$, that whole part cancels out and now we can apply our recursion to get $$I=\frac{n+1}{2}\int_{-1}^1 xP_nP_m-P_{n+1}P_m\ dx +\frac{m+1}{2}\int_{-1}^1 xP_mP_n -P_{m+1}P_n\ dx$$ Using the definition of $I$ we can rewrite all this as $$-2I + (n+1)I + (m+1)I =(n+1)\int_{-1}^1 P_{n+1}P_m\ dx + (m+1)\int_{-1}^1 P_{m+1}P_n\ dx$$ And we can simplify the integrals of $P_aP_b$ using your well known integral to get $$I(n+m) = \frac{2(n+1)}{2(n+1)+1}\delta_{n+1,m} + \frac{2(m+1)}{2(m+1)+1}\delta_{n,m+1}$$ I'll let you take it from here.