Đáp án:
$\displaystyle\int\dfrac{\sqrt{x^2 +1}}{x}dx=\sqrt{x^2 +1} + \ln\left|\dfrac{x}{\sqrt{x^2+1}} -1\right| + C$
Giải thích các bước giải:
$\displaystyle\int\dfrac{\sqrt{x^2 +1}}{x}dx$ $(*)$
Đặt $x = \tan\theta$
$\Rightarrow dx = \sec^2\theta d\theta$
$(*) \Leftrightarrow \displaystyle\int\dfrac{\sqrt{\tan^2\theta + 1}}{\tan\theta}\sec^2\theta d\theta$
$= \displaystyle\int\dfrac{\sqrt{\sec^2\theta}}{\tan\theta}\sec^2\theta d\theta$
$= \displaystyle\int\dfrac{\sec\theta.(\tan^2\theta + 1)}{\tan\theta}d\theta$
$= \displaystyle\int\sec\theta.\left(\dfrac{\tan^2\theta}{\tan\theta} + \dfrac{1}{\tan\theta}\right)d\theta$
$= \displaystyle\int\left(\sec\theta.\tan\theta + \sec\theta.\dfrac{1}{\tan\theta}\right)d\theta$
$= \displaystyle\int\left(\sec\theta.\tan\theta + \dfrac{1}{\cos\theta}\cdot\dfrac{\cos\theta}{\sin\theta}\right)d\theta$
$=\displaystyle\int\left(\sec\theta.\tan\theta + \dfrac{1}{\sin\theta}\right)d\theta$
$=\displaystyle\int\left(\sec\theta.\tan\theta +\csc\theta\right)d\theta$
$= \sec\theta + \ln\left|\tan\dfrac{\theta}{2}\right| + C$
$= \sqrt{x^2 +1} + \ln\left|\dfrac{x}{\sqrt{x^2+1}} -1\right| + C$
