Chứng minh: $\int\dfrac{1}{\sqrt{x^2+1}}dx=\ln(x+\sqrt{x^2+1})+C$
Đặt: $x=\tan t \Rightarrow dx=\dfrac{1}{\cos^2t}dt$
Khi đó ta có:
$\int\dfrac{1}{\sqrt{x^2+1}}dx$
$=\int\dfrac{1}{\sqrt{\tan^2t+1}}\dfrac{dt}{\cos^2t}$
$=\int\dfrac{dt}{\cos t}$
$=\int\dfrac{\dfrac{1}{\cos^2t}+\dfrac{\tan t}{\cos t}}{\dfrac{1}{\cos t}+\tan t}dt$
$=\int\dfrac{d(\dfrac{1}{\cos t}+\tan t)}{\dfrac{1}{\cos t}+\tan t}$
$=\ln(\dfrac{1}{\cos t}+\tan t)+C$
$=\ln(\sqrt{x^2+1}+x)+C$