Answer
$0$
Work Step by Step
Re-arrange the given integral as follows:
$\int_{-\infty}^{0} \dfrac{xdx}{(x^2+4)^{3/2}}+\int_{0}^{\infty} \dfrac{xdx}{(x^2+4)^{3/2}}=\lim\limits_{p \to -\infty}\int_{p}^{0} \dfrac{xdx}{(x^2+4)^{3/2}}+\int_{0}^{\infty} \lim\limits_{q \to \infty}\dfrac{xdx}{(x^2+4)^{3/2}}$ ....(1)
Plug $u(x)=x^2+4 \implies du=2x dx$ and $u(p)=p^2+4; u(q)=q^2+4$
Then, we have $(\dfrac{1}{2}) \int_{p^2+4}^{1} u^{-3/2} du+(\dfrac{1}{2}) \int_{1}^{q^2+4} u^{-3/2} du=[\dfrac{-1}{\sqrt u}]_{p^2+4}^{1}+[\dfrac{-1}{\sqrt u}]_{1}^{q^2+4}$
Thus, equation (1) becomes
$\lim\limits_{p \to -\infty}(\dfrac{1}{\sqrt \infty})+\lim\limits_{q \to \infty}(\dfrac{1}{\sqrt \infty})=0-0=0$
