$I=\int \dfrac{xdx}{x^2 \sqrt{x^2 +4}}$ đặt $\sqrt{x^2 +4} = t \Rightarrow x^2 =t^2 -4 \Rightarrow xdx =tdt$
$I=\int \dfrac{tdt}{(t^2-4)t}=\int \dfrac{1}{(t-2)(t+2)}dt =\dfrac{1}{4}\int \bigg (\dfrac{1}{t-2}-\dfrac{1}{t+2} \bigg ) dt$
$=\dfrac{1}{4} \bigg (\ln |t-2| -\ln|t+2| \bigg ) +C$
$=\dfrac{1}{4}\ln \bigg | \dfrac{t-2}{t+2} \bigg | + C =\dfrac{1}{4}\ln \bigg | \dfrac{\sqrt{x^2+4}-2}{\sqrt{x^2+4}+2} \bigg |+C$
tự thế cận nha