This is a generalized version of # 4618. Let $x=a\tan{t}$, then $dx=a\sec^2{t}dt$. Therefore

$$\int\frac{1}{\sqrt{x^2+a^2}}dx=\int\sec{t}dt=\ln|a\tan{t}+a\sec{t}|+C=\boxed{\ln|x+\sqrt{x^2+a^2}|+C}$$

Integral of $\sec{t}dt$ is taken from # 4617.