Đặt: $\sqrt{x^2+4}=t \Rightarrow x^2+4=t^2 \Rightarrow xdx=tdt$
Đổi cận: $x=\sqrt5 \Rightarrow t=3$
$x=2\sqrt3 \Rightarrow t=4$
Ta có:
$\int\limits_{\sqrt5}^{2\sqrt3}\dfrac{dx}{x\sqrt{x^2+4}}$
$=\int\limits_3^4\dfrac{tdt}{(t^2-4)t}$
$=\int\limits_3^4\dfrac{dt}{t^2-4}$
$=\dfrac{1}{4}\int\limits_3^4\left(\dfrac{1}{t-2}-\dfrac{1}{t+2}\right)dt$
$=\dfrac{1}{4}\ln\left|\dfrac{t-2}{t+2}\right|\left|\begin{array}{l}4\\3\end{array}\right.=\dfrac{1}{4}\ln\dfrac{5}{3}$