Answer
Diverges; $0$
Work Step by Step
Apply the limit comparison test.
$\int_{0}^{\infty} \dfrac{ 2x dx}{x^2+1}=\lim\limits_{a \to \infty} \int_{0}^{a}\dfrac{2 x dx}{x^2+1} $
$\implies \lim\limits_{a \to \infty}[\ln (x^2+1)]_{-a}^a=\lim\limits_{a \to \infty} [\ln (a^2+1) -\ln 1]$
So, the integral $\int_{0}^{\infty} \dfrac{ 2x dx}{x^2+1}$ diverges and so, the integral $\int_{-\infty}^{\infty} \dfrac{ 2x dx}{x^2+1}$ also diverges.
Now, $\lim\limits_{a \to \infty} \int_{-a}^a \dfrac{2 x dx}{x^2+1}
\\=\lim\limits_{a \to \infty}[\ln (x^2+1)]_{-a}^a \\=\lim\limits_{a \to \infty} [\ln (a^2+1) -\ln (a^2+1)] \\=\lim\limits_{a \to \infty} (0) \\=0$
