Poisson Integral relation

Question

If $$ I_n(r) = \int_0^\pi \frac{\cos nx}{r^2-2r\cos x+1} \, dx $$ How to prove that

$$ I_{n-1}(r)+I_{n+1}(r)= \left(r+\frac{1}{r}\right)I_n(r)\text{ ?}$$

I only find that $$I_{n-1}(r)+I_{n+1}(r)= \int_0^\pi \frac{2\cos nx\cos x}{r^2-2r\cos x+1} \, dx$$

Answer 1

$$\begin{aligned} I_{n-1}(r)+I_{n+1}(r) &=\int_0^{\pi}\frac{2\cos(nx) \cos x}{r^2-2r\cos x+1}\,dx \\ &=-\frac{1}{r}\int_0^{\pi} \frac{\cos(nx)(r^2-2r\cos x+1-r^2-1)}{r^2-2r\cos x+1}\,dx \\ &=-\frac{1}{r}\int_0^{\pi}\cos(nx)\,dx+\frac{r^2+1}{r}\int_0^{\pi} \frac{\cos nx}{r^2-2r\cos x+1}\,dx\\ &=\left(r+\frac{1}{r}\right)I_n(r) \\ \end{aligned}$$