Answer
\[\ln\left|\frac{\sqrt{x^2+1}-1}{x}\right|+C\]
Work Step by Step
Let \[I=\int\frac{dx}{x\sqrt{x^2+1}}\;\;\;\ldots (1)\]
Substitute $\;x=\tan \theta\;\;\;\;\ldots (2)$
\[\Rightarrow dx=\sec^2 \theta\:d \theta\]
(1) becomes
\[I=\int\frac{\sec^2 \theta}{\tan \theta\sqrt{\tan^2 \theta+1}}d \theta\]
\[I=\int\frac{\sec^2 \theta}{\tan \theta\sec \theta}d \theta\]
\[I=\int\frac{\sec \theta}{\tan \theta}d \theta\]
\[I=\int\frac{\left(\frac{1}{\cos \theta }\right)}{\left(\frac{\sin \theta }{\cos \theta}\right)}d\theta=\int\csc \theta \]
\[I=\ln|\csc\theta-\cot\theta|+C \;\;\;\ldots (3)\]
$C$ is constant of integration
From (2)
\[\tan\theta=x\]
\[\Rightarrow \cot \theta=\frac{1}{x}\;\;\;\ldots (4)\]
\[\csc\theta=\sqrt{1+\cot^2\theta}=\sqrt{1+\frac{1}{x^2}}\]
\[\csc\theta=\frac{\sqrt{x^2+1}}{x}\;\;\;\ldots (5)\]
Using (4), (5) in (3)
\[I=\ln\left|\frac{\sqrt{x^2+1}}{x}-\frac{1}{x}\right|+C\]
\[I=\ln\left|\frac{\sqrt{x^2+1}-1}{x}\right|+C\]
Hence \[I=\ln\left|\frac{\sqrt{x^2+1}-1}{x}\right|+C.\]
