Answer
$0$
Work Step by Step
Consider $f(x)= \int_{-\infty}^{\infty} \dfrac{2 x dx}{(x^2+1)^2}$
This can be re-written as: $\int_{-\infty}^{\infty} \dfrac{2 x dx}{(x^2+1)^2}=\int_{-\infty}^{0} \dfrac{2 x dx}{(x^2+1)^2}+\int_{0}^{\infty} \dfrac{2 x dx}{(x^2+1)^2}$
Let us consider $x^2+1=p(x) \implies dp=2x dx$
also, $p(k)=k^2+1$ and $p(l)=l^2+1$
This implies that $\int_{-\infty}^{0} \dfrac{2 x dx}{(x^2+1)^2}+\int_{0}^{\infty} \dfrac{2 x dx}{(x^2+1)^2}=\lim\limits_{k \to -\infty} \int_{k}^{0} \dfrac{2 x dx}{(x^2+1)^2}+\lim\limits_{l \to \infty}\int_{0}^{l} \dfrac{2 x dx}{(x^2+1)^2}$
and, $\lim\limits_{k \to -\infty} [\dfrac{dp(k)}{k^2})]_{k^2+1}^1+\lim\limits_{l \to \infty}[\dfrac{dp(l)}{l^2})]_1^{l^2+1}=\lim\limits_{k \to -\infty}[\dfrac{1}{(k)^2+1}]-1+\lim\limits_{l \to \infty}[1-\dfrac{1}{(l)^2+1}]$
Thus, $\int_{-\infty}^{\infty} \dfrac{2 x dx}{(x^2+1)^2}=-1+1=0$
