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How to find $$\displaystyle \int_0^\infty \frac{1}{(1+x^2)^n}dx\quad ?$$

For $n=1$, we have $(\arctan x) |_0^\infty = \frac{\pi}{2}.$

I tried to integrate by parts to get recurrent formula:

$$ \int_0^\infty \frac{1}{(1+x^2)^n}dx = \left. \frac{x}{(1 + x^2)^n} \right|_0^\infty ( = 0) + 2n\int_0^\infty \frac{x^2}{(x^2 + 1)^{n+1}}$$

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Hint:

$$I_n:=\int_0^\infty\frac{dx}{(x^2+1)^n}=\frac{x}{(x^2+1)^n}+2n\int_0^\infty\frac{x^2}{(x^2+1)^{n+1}}dx \\=\left.\frac{x}{(x^2+1)^n}\right|_0^\infty+2n\int_0^\infty\frac{x^2+1-1}{(x^2+1)^{n+1}}dx=2n(I_n-I_{n+1})$$

gives the recurrence

$$I_{n+1}=\frac{2n-1}{2n}I_n.$$

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