Question

Use a table of integrals with forms involving $$\sqrt{u^{2} \pm a^{2}}$$ to find $$\int \frac{\sqrt{x^{2}-9}}{3 x} d x$$

Short answer

The solution to the integral \(\int \frac{\sqrt{x^{2}-9}}{3 x} dx\) is \(\sqrt{x^2 - 9} - 3 \ln |x + \sqrt{x^2 - 9}|\).

Step-by-step solution

Recognize the standard form

Recognize the given integral \(\int \frac{\sqrt{x^{2}-9}}{3 x} dx\) as the form \(\int \frac{\sqrt{u^{2}-a^{2}}}{u} du\). In this case, \(u = x\) and \(a = 3\).

Apply the formula

From the table of integrals, we know that \(\int \frac{\sqrt{u^{2}-a^{2}}}{u} du = \sqrt{u^2 - a^2} - a \ln |u + \sqrt{u^2 - a^2}|\). Now, insert \(u = x\) and \(a = 3\) into the formula.

Simplify the result

After substitution, the integral becomes \(\sqrt{x^2 - 3^2} - 3 \ln |x + \sqrt{x^2 - 3^2}|\). This can further be simplified as \(\sqrt{x^2 - 9} - 3\ln |x + \sqrt{x^2 - 9}|\).