I am trying to find the integration: $$ \int_{0}^{1}{xP_n\left(x\right)}\,dx $$ I know that I should split Rodrigues's formula up into two parts $P_{2n}\left(x\right)$ for even terms and $P_{2n+1}\left(x\right)$ for odd terms. I also know identity formulas such as
1. $$ \int_{0}^{1}{P_{2n}\left(x\right)}\,dx=0 $$ 2. $$ \int_{0}^{1}{P_{2n}\left(x\right)}\,dx=\frac{\left(-1\right)^n\left(2n\right)!} {2^{2n+1}\left(n!\right)\left(n+1\right)!} $$ 3. $$ \left(2n+1\right)P_n=P_{n+1}^\prime-P_{n-1}^\prime $$
I know that this integral goes to $0$ for the odd terms. For the even terms I have a feeling I should be able to do integration by parts to begin but I can't seem to figure out what to use as $u,du,v,dv$... since I only know the definite integration of $P_{2n+1}$.
Is there an integration by parts rule I don't know or is this the wrong method?
You have enough knowledge for at least two ways to solve the problem.
Let $I_n=\int_0^1 xP_n(x)\,dx$. Clearly $I_0=1/2$ and $I_1=1/3$.
For odd $n>1$ we have $I_n=\frac12\int_{-1}^1 xP_n(x)\,dx=0$ by orthogonality.
From now on, assume $n>1$ is even.
Solution 1. Use $(2n+1)P_n(x)=P_{n+1}'(x)-P_{n-1}'(x)$ and integration by parts: $$(2n+1)I_n=x\big(P_{n+1}(x)-P_{n-1}(x)\big)\Bigg|_{x=0}^{x=1}-\int_0^1\big(P_{n+1}(x)-P_{n-1}(x)\big)\,dx,$$ i.e. $(2n+1)I_n=J_{n-1}-J_{n+1}$, where $J_n=\int_0^1 P_n(x)\,dx$ is found a similar way: $$(2n+1)J_n=\big(P_{n+1}(x)-P_{n-1}(x)\big)\Bigg|_{x=0}^{x=1}=P_{n-1}(0)-P_{n+1}(0),$$ and it remains to recall a formula for $P_n(0)$ and do some simplifications.
Solution 2. Use Rodrigues' formula $P_n(x)=\frac1{2^n n!}\frac{d^n}{dx^n}(x^2-1)^n$ and integration by parts: \begin{align} 2^n n!\,I_n&=\int_0^1 x\frac{d^n}{dx^n}(x^2-1)^n\,dx \\&=x\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n\Bigg|_{x=0}^{x=1}-\frac{d^{n-2}}{dx^{n-2}}(x^2-1)^n\Bigg|_{x=0}^{x=1} \\&=\frac{d^{n-2}}{dx^{n-2}}(x^2-1)^n\Bigg|_{x=0}=(n-2)!\underbrace{[x^{n-2}](x^2-1)^n}_{\text{the-coefficient-of}} \\&=(n-2)!\binom{n}{n/2-1}(-1)^{n/2+1}, \end{align} so that the answer is $I_n=\dfrac{(-1)^{n/2+1}(n-2)!}{2^n(n/2-1)!(n/2+1)!}$ (for even $n>1$).