Compute $\displaystyle \displaystyle \lim_{n\to\infty} \int_0^\infty \frac{\tan^2x}{(n+\tan^2x)(x^2+1)}dx $.
Let $\displaystyle \displaystyle I_n:= \int_0^{+\infty}\dfrac{\tan^2 x}{(n+\tan^2 x)(1+x^2)}dx$. The sequence $\displaystyle \left\{I_n\right\}_n$ is decreasing and positive so the limit exists. We have
$\displaystyle \displaystyle I_n = \int_0^{+\infty}\dfrac{\tan^2 x}{(n+\tan^2 x)(1+x^2)}dx
=\sum_{j=0}^{+\infty}\int_{j\pi}^{(j+1)\pi}\dfrac{ \tan^2 x}{(n+\tan^2 x)(1+x^2)}dx
$
and by the substitution $\displaystyle t=x-j\pi$ we get $\displaystyle \displaystyle I_n =\sum_{j=0}^{+\infty}\int_0^{\pi}\dfrac{\tan^2 t}{(n+\tan^2t)(1+(t+j\pi)^2)}dt$.
For all $\displaystyle j$ we have the inequality $\displaystyle \dfrac 1{1+((j+1)\pi)^2}\leq \dfrac 1{1+(t+j\pi)^2}\leq \dfrac 1{1+(j\pi)^2}$ hence $\displaystyle \displaystyle\int_0^{\pi}\dfrac{\tan^2 t}{n+\tan^2t}dt \sum_{j=0}^{+\infty}\dfrac 1{1+((j+1)\pi)^2}\leq I_n\leq \int_0^{\pi}\dfrac{\tan^2 t}{n+\tan^2t}dt \sum_{j=0}^{+\infty}\dfrac 1{1+(j\pi)^2}$.
Now we try to compute $\displaystyle \displaystyle J_n :=\int_0^{\pi}\dfrac{\tan^2 t}{n+\tan^2t}dt = \int_0^{\frac{\pi}2}\dfrac{\tan^2 t}{n+\tan^2t}dt +\int_0^{\frac{\pi}2}\dfrac 1{n\tan^2t+1}dt$. By putting $\displaystyle x=\tan t$ we get
$\displaystyle \displaystyle J_n =\int_0^{+\infty}\dfrac{x^2}{(n+x^2)(1+x^2)}dx+\in t_0^{+\infty}\dfrac{1}{(nx^2+1)(1+x^2)}dx$. We can see that the two integrals are the same (by the substitution $\displaystyle y=\frac 1x$). So we only have to compute the first one.
We have $\displaystyle \displaystyle \dfrac{x^2}{(n+x^2)(1+x^2)}dx =\dfrac 1{n-1}\left(\dfrac n{n+x^2}-\dfrac 1{1+x^2}\right)$. Then $\displaystyle \displaystyle J_n =2\dfrac 1{n-1}\int_0^{+\infty}\dfrac {dt}{1+\left(\dfrac t{\sqrt n}\right)^2}-2\dfrac{\pi}{2(n-1)} =\dfrac{(\sqrt n-1)\pi}{n-1}$ and we conclude the limit we were looking for is $\displaystyle 0$.