Step 1: Substitution

Let $x = a \sec{\theta}$. Then, $dx = a \sec{\theta} \tan{\theta} d\theta$.

Step 2: Simplify the Integral

Substitute $x = a \sec{\theta}$ into the integral:

$\int \frac{1}{x} \sqrt{\frac{x+a}{x-a}} dx = \int \frac{1}{a \sec{\theta}} \sqrt{\frac{a \sec{\theta} + a}{a \sec{\theta} - a}} (a \sec{\theta} \tan{\theta}) d\theta$

$= \int \frac{1}{a \sec{\theta}} \sqrt{\frac{\sec{\theta} + 1}{\sec{\theta} - 1}} (a \sec{\theta} \tan{\theta}) d\theta = \int \sqrt{\frac{\sec{\theta} + 1}{\sec{\theta} - 1}} \tan{\theta} d\theta$

Step 3: Further Simplification

Multiply the numerator and denominator inside the square root by $(\sec{\theta} + 1)$:

$\sqrt{\frac{\sec{\theta} + 1}{\sec{\theta} - 1}} = \sqrt{\frac{(\sec{\theta} + 1)^2}{\sec^2{\theta} - 1}} = \sqrt{\frac{(\sec{\theta} + 1)^2}{\tan^2{\theta}}} = \frac{\sec{\theta} + 1}{\tan{\theta}}$

So, the integral becomes:

$\int \frac{\sec{\theta} + 1}{\tan{\theta}} \tan{\theta} d\theta = \int (\sec{\theta} + 1) d\theta$

Step 4: Integrate

$\int (\sec{\theta} + 1) d\theta = \int \sec{\theta} d\theta + \int 1 d\theta = \ln|\sec{\theta} + \tan{\theta}| + \theta + C$

Step 5: Substitute Back

Since $x = a \sec{\theta}$, we have $\sec{\theta} = \frac{x}{a}$. Then $\theta = \sec^{-1}(\frac{x}{a})$.

Also, $\tan{\theta} = \sqrt{\sec^2{\theta} - 1} = \sqrt{\frac{x^2}{a^2} - 1} = \sqrt{\frac{x^2 - a^2}{a^2}} = \frac{\sqrt{x^2 - a^2}}{a}$

Substituting back, we get:

$\ln|\frac{x}{a} + \frac{\sqrt{x^2 - a^2}}{a}| + \sec^{-1}(\frac{x}{a}) + C = \ln|\frac{x + \sqrt{x^2 - a^2}}{a}| + \sec^{-1}(\frac{x}{a}) + C$

$= \ln|x + \sqrt{x^2 - a^2}| - \ln|a| + \sec^{-1}(\frac{x}{a}) + C$

Since $-\ln|a|$ is a constant, we can absorb it into the constant of integration:

$\ln|x + \sqrt{x^2 - a^2}| + \sec^{-1}(\frac{x}{a}) + C$