Đặt $t= e^{x}$,ta có:
$e^{x}dx=dt \Rightarrow dx= \frac{dt}{e^{x}}= \frac{dt}{t}$
khi $x= 0$ thì t=1, khi $x =\ln \sqrt{3}$ thì $t=\sqrt{3}$
I= $\int\limits_{1}^{\sqrt{3}}\frac{1}{t^{4} + t^{2}}dt = \int\limits_{1}^{\sqrt{3}}(\frac{1}{t^{2}}- \frac{1}{t^{2}+1})dt$
$= -\frac{1}{t} \bigg|_1^\sqrt{3} - \int\limits_{1}^{\sqrt{3}}\frac{1}{t^{2}+1}dt= \frac{3-\sqrt{3}}{3}- I_{1}$
Tính $I_{1}$.
Đặt $t= tana$, ta có:
$dt= (1+tan^{2}a)da$
khi t= 1 thì $a= \frac{\pi }{4}$, khi $t=\sqrt{3}$ thì $a= \frac{\pi }{3}$
$I_{1}= \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{3}}da = a\bigg|_\frac{\pi}{4}^\frac{\pi}{3} = \frac{\pi}{12}$
$\Rightarrow I = \frac{3-\sqrt{3}}{3} - \frac{\pi}{12}$.