Answer
$0$
Work Step by Step
Re-arrange the given integral as follows:
$\int_{-\infty}^{0} \dfrac{2xdx}{(x^2+1)^2}+\int_{0}^{\infty} \dfrac{2xdx}{(x^2+1)^2}=\lim\limits_{p \to -\infty}\int_{p}^{0} \dfrac{2xdx}{(x^2+1)^2}+\lim\limits_{q \to \infty}\int_{0}^{q} \dfrac{2xdx}{(x^2+1)^2}$ ....(1)
Plug $u(x)=x^2+1 \implies du=2x dx$ and $u(p)=p^2+1; u(q)=q^2+1$
Then, we have $\int_{p^2+1}^{1} \dfrac{du}{u^2}+\int_{1}^{c^2+1} \dfrac{du}{u^2}=[\dfrac{-1}{u}]_{p^2+1}^{1}+[\dfrac{-1}{u}]_{1}^{q^2+1}$
Thus, equation (1) becomes
$\lim\limits_{p \to -\infty}(\dfrac{1}{p^2+1}-1)+\lim\limits_{q \to \infty}(1-\dfrac{1}{q^2+1})=-1+1=0$
