Let $I_n(x)=\int^x_0\frac{1}{(1+t^2)^n}dt$ for $x>0$ and $n=1,2...$
I manipulated with the integral a bit and got the recursion: $I_{n+1}(x)=\frac{x}{2n(1+x^2)^n}+\frac{2n-1}{2n}I_n(x)$
I now need to prove that $I_n(\infty)=\int^{\infty}_0\frac{1}{(1+t^n)^n}dt=\lim_{x\to\infty}I_n(x)$ exists for all $n=1,2,...$
I prove it by induction:
Base case:
$I_1(\infty)=\int^{\infty}_0\frac{1}{1+t^2}dt=\arctan\Big|^\infty_0=\frac{\pi}{2}$
So it is true for base case. Now, I assume that it's true for the $n$'th case. I try to prove that it's true for the $n+1$'th case.
$$I_{n+1}(\infty)=\frac{x}{2n(1+x^2)^n}\Big|^{\infty}_0+\frac{2n-1}{2n}I_n(\infty)=\frac{2n-1}{2n}I_n(\infty)$$
Since $I_n(\infty)$ exists, $I_{n+1}(\infty)$ exists.
Is this correct?