calculus - How to compute this integral $int^infty_0 frac{x^{n-1}}{(1+x^2)^n}dx$?

For the integral $\int^\infty_0 \frac{x^{n-1}}{(1+x^2)^n}dx$, $n\in \mathbb{N}$, I would like to solve it in the following way:
$$
\int^\infty_0 \frac{x^{n-1}}{(1+x^2)^n}dx \overset{x=\tan t}{===} \int^{\frac{\pi }{2}}_{0}\left (\tan t \right )^{^{n-1}}dt.
$$
When $n=1$ the answer is $\frac{\pi}{2}$. However, the following equation is valid:
$$

\int^{\frac{\pi }{2}}_{0}\tan tdt=-\ln |\cos t|\mid^{\frac{\pi}{2}}_{t=0} \ .
$$

Hence, we know taht this integral can not be convergent when $n=2$. But I have proved that this integral is convergent for $n\in \mathbb{N}$.

Question:

$1.$ Where I went wrong?

$2.$ if it is truly convergent how to compute this integral for $n\in \mathbb{N}$?

Answer

If we put $x=\tan t$, then $dx=\sec^2 t\,dt$. The bottom becomes $\sec^{2n} t$. So we end up integrating $\frac{\tan^{n-1}t}{\sec^{2n-2} t}$, that is, $\sin^{n-1} t \cos^{n-1}t$. One way to continue is to rewrite in terms of $\sin 2t$, and use a Reduction Formula. One can also obtain a reduction formula directly for the original function, without travelling through the trigonometric substitution process.