Identify the form of the integral.

This integral is of the form $\int \sqrt{x^2 - a^2} \, dx$, which suggests using a trigonometric substitution.

For expressions involving $\sqrt{x^2 - a^2}$, the standard substitution is $x = a \sec \theta$.
From this substitution, we can find $dx$:

Also, we can express $\sqrt{x^2 - a^2}$ in terms of $\theta$:

Assuming $x > a$ (or $x < -a$) and $\theta$ is in an appropriate range where $\tan \theta \ge 0$, we have $\sqrt{x^2 - a^2} = a \tan \theta$.

We use the known integral formulas:

Substitute these back into the expression:

From $x = a \sec \theta$, we have $\sec \theta = \frac{x}{a}$.
To find $\tan \theta$, we use the identity $\tan \theta = \sqrt{\sec^2 \theta - 1}$:

Now substitute these expressions back into the result:

Distribute $\frac{a^2}{2}$:

Using the logarithm property $\ln\left|\frac{A}{B}\right| = \ln|A| - \ln|B|$:

Since $\frac{a^2}{2} \ln|a|$ is a constant, it can be absorbed into the arbitrary constant $C$.

The final answer is $\boxed{\frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln|x + \sqrt{x^2 - a^2}| + C}$.