Recurrence relation for $\int_{0}^{\infty} \frac{1}{(1+x^2/a)^n}dx$

$$I_n=\int_{0}^{\infty} \frac{1+x^2/a}{(1+x^2/a)^{n+1}}dx=\underbrace{\int_{0}^{\infty} \frac{1}{(1+x^2/a)^{n+1}}dx}_{=I_{n+1}}+\frac12\int_0^\infty \frac{2x^2/a}{(1+x^2/a)^{n+1}}dx$$ Now since: $$\int \frac{2x/a}{(1+x^2/a)^{n+1}}dx=\int \frac{(1+x^2/a)'}{(1+x^2/a)^{n+1}}dx=-\frac{1}{n} \frac{1}{(1+x^2/a)^n}+C$$ $$\Rightarrow\int_0^\infty \frac{2x^2/a}{(1+x^2/a)^{n+1}}dx=\int_0^\infty x\left(-\frac{1}{n}\frac{1}{(1+x^2/a)^{n}}\right)'dx$$ $$=-\underbrace{\frac{x}{n}\left(\frac{1}{\left(1+x^2/a\right)^n}\right)\bigg|_0^\infty}_{=0}+\frac{1}{n}\int_0^\infty \frac{1}{(1+x^2/a)^n}dx=\frac{1}{n} I_n$$ $$\Rightarrow I_n=I_{n+1}+\frac{1}{2n} I_n\Rightarrow I_{n+1}=\frac{2n-1}{2n}I_n$$

Hint:

Calculate the integral $I_{\color{red}n,a}$ by parts, setting \begin{align}&&u&=\frac 1{\biggl(1+\cfrac{x^2}{a}\biggr)^n},\qquad &\mathrm d v&=\mathrm dx,&\qquad&\\ &\text{whence}\qquad&\qquad\mathrm du&=-\frac n{\biggl(1+\cfrac{x^2}{a}\biggr)^{n+1}}\frac{2x}{a},&& v=x,\\ &\text{which yields } &I_{n,a}&=\begin{array}{c|}\dfrac x{\biggl(1+\tfrac{x^2}{a}\biggr)^n}\end{array}_0^\infty+2n\int_0^\infty\rlap{\dfrac{\tfrac{x^2}a}{\biggl(1+\tfrac{x^2}{a}\biggr)^{\!n}}\,\mathrm dx}. \end{align} Can you end the computation?