Integrals with Legendre polynomials I'm doing exercise 2.5 in Differential Equations by King, Billingham, Otto and the problem states 

Establish the results

a) $\int_{-1}^1{xP_n(x)P_{n-1}dx} = \frac{2n}{4n^2-1}$
b) $\int_{-1}^1{P_n(x)P_{n+1}'(x)dx} = 2$
c) $\int_{-1}^1{xP_n'(x)P_{n}dx} = \frac{2n}{2n+1}$


The furthest I've reached is on question b) where I got:
$2-\int_{-1}^1{P_n'(x)P_{n+1}dx}$ by doing integration by parts.
Can someone give me some hints on how to solve these or where to start?
Thank you
 A: Here is an answer for a.  Recall that
$$(n+1)P_{n+1}(x)=(2n+1)xP_n(x)-nP_{n-1}(x).$$
So $xP_n(x)=\frac{n+1}{2n+1}P_{n+1}(x)+\frac{n}{2n+1}P_{n-1}(x)$.  Plug this into LHS to get
$$\int_{-1}^1xP_n(x)P_{n-1}(x)dx=\frac{n+1}{2n+1}\int_{-1}^{1}P_{n+1}(x)P_{n-1}(x)dx+\frac{n}{2n+1}\int_{-1}^1P_{n-1}^2(x)dx.$$
Since  $\int_{-1}^1 P_m(x)P_n(x)dx=\frac{2}{2n+1}\delta_{m,n}$, the first term is $0$ while the second term is $$\frac{n}{2n+1}\left(\frac{2}{2(n-1)+1}\right)=\frac{2n}{4n^2-1}.$$
Here is an answer for b.  LHS equals
$$\int_{-1}^1P_n(x)dP_{n+1}(x)=P_n(x)P_{n+1}(x)\Big|_{-1}^1-\int_{-1}^1P_n'(x)P_{n+1}(x)dx$$
but $P_n(1)=P_{n+1}(1)=1$ while $\left\{P_n(-1),P_{n+1}(-1)\right\}=\{-1,1\}$ so
$$\int_{-1}^1P_n(x)P'_{n+1}(x)dx=\int_{-1}^1P_n(x)dP_{n+1}(x)=2-\int_{-1}^1P_n'(x)P_{n+1}(x)dx.$$
But $P_0,P_1,\ldots,P_{n-1}$ span the space of polynomials degree at most $n-1$ and $\deg P'_{n}=n-1$, so $P'_n$ is a linear combination of $P_0,P_1,\ldots,P_{n-1}$.  By orthogonality relations of Legendre polynomials, $$\int_{-1}^1P_n'(x)P_{n+1}(x)dx=0,$$ and this proves b.
Here is an answer for c.  LHS equals
$$\frac12\int_{-1}^1 x\ d\big(P_n(x)\big)^2=\frac12\left(x\big(P_n (x)\big)^2\Big|_{-1}^1-\int_{-1}^1\big(P_n(x)\big)^2dx\right).$$
Since $P_n(\pm 1)=\pm1$ and $\int_{-1}^1 P_m(x)P_n(x)dx=\frac{2}{2n+1}\delta_{m,n}$, we see that
$$\int_{-1}^1 xP'_n(x)P_n(x)dx=\frac12\int_{-1}^1xd\big(P_n(x)\big)^2=\frac12\left(2-\frac{2}{2n+1}\right)=\frac{2n}{2n+1}.$$
