Answer
$0$
Work Step by Step
We have: $\int_{0}^{\infty} \dfrac{ 2x dx}{x^2+1}=\lim\limits_{a \to \infty} \int_{0}^{a}\dfrac{2 x dx}{x^2+1} =\lim\limits_{a \to \infty}[\ln (x^2+1)]_{-a}^a=\lim\limits_{a \to \infty} [\ln (a^2+1) -\ln 1]$
This implies that $\int_{0}^{\infty} \dfrac{ 2x dx}{x^2+1}$ diverges and so, $\int_{-\infty}^{\infty} \dfrac{ 2x dx}{x^2+1}$ diverges.
Now, $\lim\limits_{a \to \infty} \int_{-a}^a \dfrac{2 x dx}{x^2+1} =\lim\limits_{a \to \infty}[\ln (x^2+1)]_{-a}^a =\lim\limits_{a \to \infty} [\ln (a^2+1) -\ln (a^2+1)]=\lim\limits_{a \to \infty} (0)=0$
