Answer
$\displaystyle \ln|\frac{\sqrt{x^{2}+1}-1}{x}|+C$
Work Step by Step
There is a $\sqrt{x^{2}+a^{2}}$ term in the integrand, so we try the trigonometric substitution
$\displaystyle \int\frac{dx}{x\sqrt{x^{2}+1}}=\left[\begin{array}{ll}
x=\tan t & \\
dx=\sec^{2}tdt &
\end{array}\right],\displaystyle \qquad \left[\begin{array}{l}
\sec^{2}t=\tan^{2}t+1\\
\sec t=\sqrt{x^{2}+1}
\end{array}\right]$
$=\displaystyle \int\frac{\sec^{2}tdt}{\tan t\sec t}\qquad $... simplify (cancel one sec$...)$
$=\displaystyle \int\frac{\sec t}{\tan t}dt$
$=\displaystyle \int\csc tdt$
$=\ln|\csc t-\cot t|+C$
... bring back x
$\left[\begin{array}{ll}
\csc t=\dfrac{\sec t}{\tan t}=\dfrac{\sqrt{x^{2}+1}}{x}, & \cot t=\dfrac{1}{x}\\
&
\end{array}\right]$
$=\displaystyle \ln|\frac{\sqrt{x^{2}+1}}{x}-\frac{1}{x}|+C$
$=\displaystyle \ln|\frac{\sqrt{x^{2}+1}-1}{x}|+C$
