(To start with, it's perhaps easiest to think of the $P$ as Legendre polynomials, i.e., $\alpha=\beta=0$.)
Given two successive Jacobi polynomials $P_n^{(\alpha, \beta)}$, $P_{n-1}^{(\alpha, \beta)}$, it is known that $$ \int_{-1}^{+1} w_{\alpha, \beta}(x) \cdot P_n^{(\alpha, \beta)}(x) \cdot P_{n-1}^{(\alpha, \beta)}(x) \;\text{d}x = 0 $$ with $w_{\alpha, \beta}(x) = (1-x)^{\alpha} (1+x)^{\beta}$ being the weight function. (In fact this holds for any pair of $P_i$, $P_j$, $i\neq j$.)
Let's now multiply the integrand by $x$; does $$ \int_{-1}^{+1} w_{\alpha, \beta}(x) \cdot x\cdot P_n^{(\alpha, \beta)}(x) \cdot P_{n-1}^{(\alpha, \beta)}(x) \;\text{d}x $$ have an explicit representation in terms of $n, \alpha, \beta$?
Edit:
For $\alpha=\beta=0$ (i.e., Legendre polynomials), it appears from trying out different values of $n$ that $$ \int_{-1}^{+1} x\cdot P_n^{(0, 0)}(x) \cdot P_{n-1}^{(0, 0)}(x) \;\text{d}x = \frac{2n}{(2n-1) (2n+1)} $$
Due to orthogonality, we have
$$\int_{-1}^1 w_{\alpha, \beta}(x)\, x^k\, P_n^{(\alpha, \beta)}(x)\;\mathrm dx = 0;\quad k<n$$
so it suffices to consider
$$\frac1{2^{n-1}}\binom{2n+\alpha+\beta-2}{n-1}\int_{-1}^1 w_{\alpha, \beta}(x)\, x^n\, P_n^{(\alpha, \beta)}(x)\;\mathrm dx = 0$$
since the lower order terms drop out. (Note that $\dfrac1{2^n}\dbinom{2n+\alpha+\beta}{n}$ is the leading coefficient of $P_n^{(\alpha, \beta)}(x)$.)
Recall also that $x^n$ can be expressed as a series in terms of $P_n^{(\alpha, \beta)}(x)$:
$$x^n=\frac{2^n}{\binom{2n+\alpha+\beta}{n}}P_n^{(\alpha, \beta)}(x)+\cdots$$
where we again ignore lower order terms due to orthogonality. Recalling also that
$$\int_{-1}^1 w_{\alpha, \beta}(x)\, \left(P_n^{(\alpha, \beta)}(x)\right)^2\,\mathrm dx =\frac{2^{\alpha+\beta+1}\Gamma\left(n+\alpha+1\right)\Gamma\left(n+\beta+1 \right)}{(2n+\alpha+\beta+1)\Gamma\left(n+\alpha+\beta+1\right)n!}$$
we finally have
$$\int_{-1}^1 w_{\alpha, \beta}(x)\, x\, P_n^{(\alpha, \beta)}(x) \, P_{n-1}^{(\alpha, \beta)}(x) \,\mathrm dx=\frac{2^{\alpha +\beta +2}\, \Gamma (n+\alpha +1) \Gamma (n+\beta +1)}{(\alpha +\beta +2 n-1) (\alpha +\beta +2 n) (\alpha +\beta +2 n+1) \Gamma (n+\alpha +\beta)(n-1)!}$$