Tinh nguyen ham cua
$I=\int\limits_{}^{}\frac{1}{\sqrt{x^2+1}}dx$
Dat $x=tant t\in [\frac{-\pi}{2};\frac{\pi}{2}]$
$I=\int\limits_{}^{}\frac{1}{cost}dt=\int\limits_{}^{}\frac{d(sint)}{(1-sint)(1+sint)}=\frac{1}{2}\int\limits_{}^{}\frac{d(sint)}{1-sint}+\frac{1}{2}\int\limits_{}^{}\frac{d(sint)}{sint+1}=-\frac{1}{2}ln\left| {1-sint} \right|+\frac{1}{2}ln\left| {1+sint} \right|+C$
$x=\sqrt{3}=>sint=\frac{\sqrt{3}}{2}$
$x=\sqrt{8}=>sint=\frac{2\sqrt{2}}{3}$
Thay can tinh ..............