How to complete this proof of $\int_{0}^ \infty \frac{nx \arctan(x)}{(1+x)(n^2+x^2)}dx =\frac{\pi^2}{4}$?

I saw this problem:$\int_{0}^ \infty \frac{nx \arctan(x)}{(1+x)(n^2+x^2)}dx =\frac{\pi^2}{4}$ and I tried to solve it.

Here is my attempt

$\textbf{Claim: }$ If $f$ is continuous at $[0,1]$ then $\displaystyle \lim_{n \to \infty }\int_{0}^{1} \frac{nf(x)}{1+n^2x^2}dx=\frac{\pi f(0)}{2} $

Proof

Since $f $ is continuous at $[0,1]$ then $\forall \epsilon >0, \ \exists \delta> {0} $ such that if $|x|<\delta \ , |f(x)- f(0) |<\epsilon $. Since Continuous functions attend both maximum and minimum value on compact sets, let $M = 2\sup_{x\in [0,1]} |f(x) |$

$$\int_{0}^{1} \frac{nf(x)}{1+n^2x^2}dx= \int_{0}^{\delta} \frac{nf(x)}{1+n^2x^2}dx+\int_{\delta }^{1} \frac{n(f(x))}{1+n^2x^2}dx $$$$= \int_{0}^{\delta} \frac{n(f(x)- f(0))}{1+n^2x^2}dx+ \int_{0}^{\delta} \frac{nf(0)}{1+n^2x^2}dx +\int_{\delta }^{1} \frac{n(f(x))}{1+n^2x^2}dx $$

$$=f(0) \arctan(n\delta)+ \int_{0}^{\delta} \frac{n(f(x)- f(0))}{1+n^2x^2}dx \int_{\delta }^{1} \frac{n(f(x))}{1+n^2x^2}dx $$

Since $ \displaystyle \lim_{x \to \infty } \arctan(x) =\frac{\pi}{2}$ then for $\epsilon>0, \ \exists \alpha \in \mathbb{R} $ such that if $ x >\alpha $ then $|\arctan(x)- \frac{\pi}{2}| < \epsilon$ choose $N$ such that $N\delta >\alpha$ for all $n \ge N $

$$\bigg|\frac{\pi f(0)}{2}-f(0) \arctan(n\delta)- \int_{0}^{\delta} \frac{n(f(x)- f(0))}{1+n^2x^2}dx- \int_{\delta }^{1} \frac{n(f(x))}{1+n^2x^2}dx \bigg|$$

$$\le 2M\left|\frac{\pi }{2}- \arctan(n\delta) \right|+\epsilon \arctan(n\delta) < 2M\epsilon + \epsilon^2 $$

Which implies that$\displaystyle \lim_{n \to \infty }\int_{0}^{1} \frac{nf(x)}{1+n^2x^2}dx=\frac{\pi f(0)}{2} $

$$\int_{0}^ \infty \frac{nx \arctan(x)}{(1+x)(n^2+x^2)}dx= \int_{0}^ 1 \frac{nx \arctan(x)}{(1+x)(n^2+x^2)}dx+\int_{1}^ \infty \frac{nx \arctan(x)}{(1+x)(n^2+x^2)}dx$$

Since $\displaystyle \lim_{n \to \infty } \int_{0}^ 1 \frac{nx \arctan(x)}{(1+x)(n^2+x^2)}dx<\frac{\pi}{4} \lim_{n \to \infty }\int_{0}^ 1 \frac{x}{n(1+x)} dx =0 $

So $\displaystyle \lim_{n \to \infty } \int_{0}^ \infty \frac{nx \arctan(x)}{(1+x)(n^2+x^2)}dx =\lim_{n \to \infty } \int_{1}^ \infty \frac{nx \arctan(x)}{(1+x)(n^2+x^2)}dx $

Let $x=\frac{1}{t} $, $\color{red}{\text{define $\arctan(\frac{1}0) =\frac{\pi}{2}$}}$

$$\lim_{n \to \infty } \int_{1}^ \infty \frac{nx \arctan(x)}{(1+x)(n^2+x^2)}dx= \lim_{n \to \infty } \int_{0}^ 1 \frac{n \arctan(\frac{1}{t}) }{(1+t)(t^2n^2+1)}dt $$

By the above claim $$\int_{0}^ \infty \frac{nx \arctan(x)}{(1+x)(n^2+x^2)}dx =\frac{\pi^2}{4}$$

I highlighted the problem in red, I need to define $\frac{1}{0}$ and then define $\arctan \frac 1 0$ and of course this is not a good way to complete this proof, How can it be fixed ?