The integral presented is:

$\int \frac{1}{x^2 \sqrt{9x^2 - 1}} \, dx$

Solution Strategy:

We will apply a substitution to simplify the integrand. Since the term under the square root suggests a trigonometric substitution, we will use the following:

Substitution:

Let $x = \frac{1}{3} \sec \theta$, which implies: $dx = \frac{1}{3} \sec \theta \tan \theta \, d\theta$

This substitution simplifies the term $9x^2 - 1$ inside the square root: $9x^2 - 1 = 9 \left( \frac{1}{9} \sec^2 \theta \right) - 1 = \sec^2 \theta - 1 = \tan^2 \theta$ Thus, $\sqrt{9x^2 - 1} = \tan \theta$.

Now, let's rewrite the integrand in terms of $\theta$.

Working it out step-by-step:

1. After substitution, the integral becomes: $\int \frac{1}{\left( \frac{1}{9} \sec^2 \theta \right) \cdot \tan \theta} \cdot \frac{1}{3} \sec \theta \tan \theta \, d\theta$

2. Simplify the expression inside the integral.

Let me compute the details.After simplifying the expression and integrating, the result is:

$3 \sin \theta$

Step to revert the substitution:

We initially substituted $x = \frac{1}{3} \sec \theta$, which means $\sec \theta = 3x$. Therefore, $\theta = \sec^{-1}(3x)$ and we also know that:

$\sin \theta = \frac{\sqrt{9x^2 - 1}}{3x}$

Thus, the final answer becomes:

$\int \frac{1}{x^2 \sqrt{9x^2 - 1}} \, dx = \frac{\sqrt{9x^2 - 1}}{x}$

Would you like any further details or clarifications on this solution?

Related questions:

1. How does trigonometric substitution simplify integrals with square roots?
2. What are other common trigonometric substitutions for different integrals?
3. Can we apply hyperbolic substitution for integrals like this one?
4. How would you solve this integral using integration by parts?
5. Can numerical methods be used for integrals when symbolic methods are difficult?

Tip:

For integrals involving $\sqrt{a^2 x^2 - b^2}$, trigonometric substitutions like $x = \frac{b}{a} \sec \theta$ are very useful for simplifying the expression.