Answer
$$\displaystyle{\int\frac{\sqrt{1+x^2}}{x}dx=-\ln\left|\frac{\sqrt{1+x^2}}{x}+\frac{1}{x}\right|+\sqrt{1+x^2}+C}\\$$
Work Step by Step
$\displaystyle{I=\int\frac{\sqrt{1+x^2}}{x}dx}\\$
$\displaystyle \left[\begin{array}{ll} x=\tan\theta & x^2=\tan^2\theta \\ & \\ \frac{dx}{d\theta}=2\sec^2\theta & dx=2\sec^2\theta\ d\theta \end{array}\right]$
Integration by substitution
$\displaystyle{I=\int\frac{\sqrt{1+\tan^2\theta}}{\tan\theta}\sec^2\theta\ d\theta}\\
\displaystyle{I=\int\frac{\sec\theta}{\tan\theta}\sec^2\theta\ d\theta}\\
\displaystyle{I=\int cosec\theta\sec^2\theta\ d\theta}\\
\displaystyle{I=\int cosec\theta+\left(\tan^2\theta+1\right)\ d\theta}\\
\displaystyle{I=\int cosec\theta+\tan\theta\sec\theta\ d\theta}\\
\displaystyle{I=-\ln\left|cosec\theta+\cot\theta\right|+\sec\theta+C}\\$
$\sec\theta=\sqrt{1+x^2}\\
\cot\theta=\frac{1}{x}\\
cosec\theta=\frac{\sqrt{1+x^2}}{x}\\$
$\displaystyle{I=-\ln\left|\frac{\sqrt{1+x^2}}{x}+\frac{1}{x}\right|+\sqrt{1+x^2}+C}\\
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