+Đặt $\sqrt{x^2-4}=t\rightarrow t^2=x^2-4\rightarrow 2tdt=2xdx$-Đổi cận $x 2 \frac{4}{\sqrt3}$ $t 0 \frac{2}{\sqrt3}$-Đổi biến$\frac{\sqrt{x^2-4}}{x^3}dx=\frac{t^2dt}{(t^2+4)^2}=\frac{dt}{t^2+4}-\frac4{(t^2+4)^2}dt$$I=\int\limits_{0}^{\frac2{\sqrt3}}(\frac1{t^2+4}-\frac4{(t^2+4)^2})dt$+Đặt $t=2tan\alpha \alpha\in(-\frac{\pi}2;\frac{\pi}2)$$\rightarrow dt=\frac{2d\alpha}{cos^2\alpha}$-Đổi cận $t 0 \frac2{\sqrt3}$ $\alpha 0 \frac{\pi}6$$I=\int\limits_{0}^{\frac{\pi}6}(\frac{cos^2\alpha}4-4\frac{cos^4\alpha}{16}).\frac{2d\alpha}{cos^2\alpha}=(\frac{\alpha}4-\frac{sin2\alpha}8)|^{\pi/6}_0=\pi/24-\sqrt3/16$
+Đặt $\sqrt{x^2-4}=t\rightarrow t^2=x^2-4\rightarrow 2tdt=2xdx$-Đổi cận $x 2 \frac{4}{\sqrt3}$ $t 0 \frac{2}{\sqrt3}$-Đổi biến$\frac{\sqrt{x^2-4}}{x^3}dx=\frac{t^2dt}{(t^2+4)^2}=\frac{dt}{t^2+4}-\frac4{(t^2+4)^2}dt$$I=\int\limits_{0}^{\frac2{\sqrt3}}(\frac1{t^2+4}-\frac4{(t^2+4)^2})dt$+Đặt $t=2tan\alpha \alpha\in(-\frac{\pi}2;\frac{\pi}2)$$\rightarrow dt=\frac{2d\alpha}{cos^2\alpha}$-Đổi cận $t 0 \frac2{\sqrt3}$ $\alpha 0 \frac{\pi}6$$I=\int\limits_{0}^{\frac{\pi}6}(\frac{cos^2\alpha}4-\frac{cos^4\alpha}{16}).\frac{2d\alpha}{cos^2\alpha}$
+Đặt $\sqrt{x^2-4}=t\rightarrow t^2=x^2-4\rightarrow 2tdt=2xdx$-Đổi cận $x 2 \frac{4}{\sqrt3}$ $t 0 \frac{2}{\sqrt3}$-Đổi biến$\frac{\sqrt{x^2-4}}{x^3}dx=\frac{t^2dt}{(t^2+4)^2}=\frac{dt}{t^2+4}-\frac4{(t^2+4)^2}dt$$I=\int\limits_{0}^{\frac2{\sqrt3}}(\frac1{t^2+4}-\frac4{(t^2+4)^2})dt$+Đặt $t=2tan\alpha \alpha\in(-\frac{\pi}2;\frac{\pi}2)$$\rightarrow dt=\frac{2d\alpha}{cos^2\alpha}$-Đổi cận $t 0 \frac2{\sqrt3}$ $\alpha 0 \frac{\pi}6$$I=\int\limits_{0}^{\frac{\pi}6}(\frac{cos^2\alpha}4-
4\frac{cos^4\alpha}{16}).\frac{2d\alpha}{cos^2\alpha}
=(\frac{\alpha}4-\frac{sin2\alpha}8)|^{\pi/6}_0=\pi/24-\sqrt3/16$