Answer
$0$
Work Step by Step
Let $f(x)= \int_{-\infty}^{\infty} \dfrac{ x dx}{(x^2+4)^{3/2}}$
This can be re-written as:$ \int_{-\infty}^{\infty} \dfrac{ x dx}{(x^2+4)^{3/2}}=\lim\limits_{k \to \infty} \lim\limits_{l \to \infty} \int_{-k}^l\dfrac{ x dx}{(x^2+4)^{3/2}}=(\dfrac{1}{2})\lim\limits_{k \to \infty} \lim\limits_{l \to \infty} \int_{-k}^l\dfrac{ 2x dx}{(x^2+4)^{3/2}}$
Plug $x^2+4=p $ and $2x dx=dp$
This implies that $(\dfrac{1}{2}) \lim\limits_{k \to \infty} \lim\limits_{l \to -\infty} \int_{k^2+4}^{l^2+4} p^{(-3/2)} dp= \lim\limits_{k \to \infty} \lim\limits_{l \to -\infty} [(-1/\sqrt p)]_{k^2+4}^{l^2+4}=\lim\limits_{k \to \infty}\dfrac{1}{\sqrt{(k^2+4)}}-\lim\limits_{l \to \infty}\dfrac{1}{\sqrt{(l^2+4)}}$
Thus, $\lim\limits_{k \to \infty}\dfrac{1}{\sqrt{\infty}}-\lim\limits_{l \to \infty}\dfrac{1}{\sqrt{\infty}}=0$
