91Ó°ÊÓ

We notice that the denominator of the given integral can be factored into \((3x)^2 - (2)^2\). This looks like a difference of squares, and we can, therefore, make use of a hyperbolic function substitution. We'll use the hyperbolic tangent function, since the integral contains the difference of squares. Let's set: $$ 3x = 2\cosh{u} $$ Now, we find the derivative with respect to x to allow for substitution: $$ \frac{d(3x)}{dx} = 3 = \frac{d(2\cosh{u})}{dx} = 2\sinh{u}\frac{du}{dx} $$ We can see that: $$ \frac{du}{dx} = \frac{3}{2\sinh{u}} $$

Now we substitute \(3x = 2\cosh{u}\) and \(\frac{du}{dx} = \frac{3}{2\sinh{u}}\) into the integral expression: $$ \int \frac{1}{9x^2 - 4} dx = \int \frac{1}{(2\cosh{u})^2-4} \left(\frac{3}{2\sinh{u}}\right) du $$

Next, we simplify the integrand expression: $$ \int \frac{1}{(2\cosh{u})^2-4} \left(\frac{3}{2\sinh{u}}\right) du = \int \frac{3}{(4\cosh^2{u}-4)(2\sinh{u})} du $$ Recall the hyperbolic trigonometric identity: $$ \cosh^2{u} - \sinh^2{u} = 1 $$ We can rewrite the integrand as: $$ \int \frac{3}{(4(\cosh^2{u}-\sinh^2{u}))(2\sinh{u})} du = \int \frac{1}{2\sinh{u}} du $$

Integrating the simplified expression with respect to u, we get: $$ \int \frac{1}{2\sinh{u}} du = \frac{1}{2}\int\frac{1}{\sinh{u}} du = \frac{1}{2}\log(\sinh{u}+\cosh{u}) + C $$ where C is the constant of integration. Finally, we need to substitute back for x. Recall that \(3x = 2\cosh{u}\), so: $$ \cosh{u} = \frac{3x}{2} $$ $$ \sinh{u} = \sqrt{\cosh^2{u} - 1} = \sqrt{\left(\frac{3x}{2}\right)^2 - 1} $$ We back substitute to get the final answer: $$ \frac{1}{2}\log\left(\sinh{u} + \cosh{u}\right) + C = \frac{1}{2}\log\left(\sqrt{\left(\frac{3x}{2}\right)^2 - 1} + \frac{3x}{2}\right) + C $$ So, the evaluated integral is: $$ \int \frac{1}{9x^2 - 4} dx = \frac{1}{2}\log\left(\sqrt{\left(\frac{3x}{2}\right)^2 - 1} + \frac{3x}{2}\right) + C $$