Recurrence relation for $\int_{0}^{\infty} \frac{1}{(1+x^2/a)^n}dx$ 
Let$$I_{n,a} = \int_{0}^{\infty} \frac{1}{(1+x^2/a)^n}dx$$ where $a>0$.
  Show that $$I_{n+1,a} = \frac{2n-1}{2n}I_{n,a}$$

I have tried integrating by parts but it didn't work for me, and I don't know what else can I try. Can anyone please help?
 A: $$I_n=\int_{0}^{\infty} \frac{1+x^2/a}{(1+x^2/a)^{n+1}}dx=\underbrace{\int_{0}^{\infty} \frac{1}{(1+x^2/a)^{n+1}}dx}_{=I_{n+1}}+\frac12\int_0^\infty \frac{2x^2/a}{(1+x^2/a)^{n+1}}dx$$
Now since: $$\int \frac{2x/a}{(1+x^2/a)^{n+1}}dx=\int \frac{(1+x^2/a)'}{(1+x^2/a)^{n+1}}dx=-\frac{1}{n} \frac{1}{(1+x^2/a)^n}+C$$
$$\Rightarrow\int_0^\infty \frac{2x^2/a}{(1+x^2/a)^{n+1}}dx=\int_0^\infty x\left(-\frac{1}{n}\frac{1}{(1+x^2/a)^{n}}\right)'dx$$
$$=-\underbrace{\frac{x}{n}\left(\frac{1}{\left(1+x^2/a\right)^n}\right)\bigg|_0^\infty}_{=0}+\frac{1}{n}\int_0^\infty \frac{1}{(1+x^2/a)^n}dx=\frac{1}{n} I_n$$
$$\Rightarrow I_n=I_{n+1}+\frac{1}{2n} I_n\Rightarrow I_{n+1}=\frac{2n-1}{2n}I_n$$
A: 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?
A: Just in case you're interested you can solve this for any $n  \in \mathbb{R}^+$, $n \geq 1$ using the Beta and by extension the Gamma Function. 
Here we will address your integral:
\begin{equation}
 I(a,n) = \int_0^\infty \frac{1}{\left(1 +  \frac{x^2}{a}\right)^n}\:dx
\end{equation}
We begin by making the substitution $x = \sqrt{a}t$:
\begin{equation}
 I(a,n) = \int_0^\infty \frac{1}{\left(1 +  \frac{at^2}{a}\right)^n} \cdot \sqrt{a}\:dt = \sqrt{a}\int_0^\infty \frac{1}{\left(1 + t^2\right)^n}\:dt
\end{equation}
We now let $t = \tan(\theta)$:
\begin{align}
 I(a,n) &= \sqrt{a}\int_0^{\frac{\pi}{2}} \frac{1}{\left(1 + \tan^2(\theta)\right)^n} \cdot \sec^2(\theta)\:d\theta \nonumber \\
&= \sqrt{a}\int_0^{\frac{\pi}{2}} \cos^{2n - 2}(\theta)\:d\theta
\end{align}
Now from here you could form your recurrence relationship by IBP twice, but alternatively you can call upon one of the common definitions of the Beta Function:
\begin{equation}
B(a,b) = 2\int_0^{\frac{\pi}{2}} \cos^{2a - 1}(x)\sin^{2b - 1}(x)\:dx
\end{equation}
For $I(a,n)$ we observe that (1) $2a - 1 = 2n - 2$ and so  $a = \frac{2n - 1}{2}$ and (2) $2b - 1 = 0$ and so $b = \frac{1}{2}$.We also observe that we need to scale the integral by $\frac{1}{2}$:
\begin{equation}
 I(a,n) = \sqrt{a}\int_0^{\frac{\pi}{2}} \cos^{2n - 2}(\theta)\:d\theta = \frac{\sqrt{a}}{2}B\left( \frac{2n - 1}{2}, \frac{1}{2} \right)
\end{equation}
Using the relationship between the Beta and Gamma Function, this reduces to:
\begin{equation}
 I(a,n) = \frac{\sqrt{a}}{2}\cdot \frac{\Gamma\left(\frac{2n - 1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{2n - 1}{2} + \frac{1}{2} \right)} = \frac{\sqrt{a}\sqrt{\pi}}{2} \frac{\Gamma\left(\frac{2n - 1}{2}\right)}{\Gamma(n)}
\end{equation}
We can also see that this satisfies the recurrence relationship you sought. 
