The integral you've provided is:

$\int \frac{dx}{\sqrt{9x^2 + 12x + 2}}$

To solve this integral, we can use a combination of completing the square and trigonometric substitution.

Step 1: Complete the square

First, let's complete the square for the expression inside the square root.

Given: $9x^2 + 12x + 2$

Factor out the 9 from the first two terms:

$9\left(x^2 + \frac{4}{3}x\right) + 2$

To complete the square, take half the coefficient of $x$, square it, and add/subtract it inside the expression:

$9\left(x^2 + \frac{4}{3}x + \left(\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2\right) + 2 = 9\left(\left(x + \frac{2}{3}\right)^2 - \frac{4}{9}\right) + 2$

Simplifying further:

$9\left(x + \frac{2}{3}\right)^2 - 4 + 2 = 9\left(x + \frac{2}{3}\right)^2 - 2$

So, the expression inside the square root becomes:

$\sqrt{9\left(x + \frac{2}{3}\right)^2 - 2}$

Step 2: Use trigonometric substitution

Now, set:

$u = x + \frac{2}{3}$ $du = dx$

The integral becomes:

$\int \frac{du}{\sqrt{9u^2 - 2}}$

This is a standard form that can be handled using trigonometric substitution. Let:

$u = \frac{\sqrt{2}}{3} \sec(\theta)$

Then:

$du = \frac{\sqrt{2}}{3} \sec(\theta) \tan(\theta) d\theta$

Substituting back into the integral:

$\int \frac{\frac{\sqrt{2}}{3} \sec(\theta) \tan(\theta) d\theta}{\sqrt{9 \left(\frac{2}{9} \sec^2(\theta)\right) - 2}} = \int \frac{\frac{\sqrt{2}}{3} \sec(\theta) \tan(\theta) d\theta}{\sqrt{2\sec^2(\theta) - 2}}$

Simplifying further:

$\int \frac{\frac{\sqrt{2}}{3} \sec(\theta) \tan(\theta) d\theta}{\sqrt{2(\sec^2(\theta) - 1)}} = \int \frac{\frac{\sqrt{2}}{3} \sec(\theta) \tan(\theta) d\theta}{\sqrt{2\tan^2(\theta)}} = \int \frac{\frac{\sqrt{2}}{3} \sec(\theta) \tan(\theta) d\theta}{\sqrt{2}\tan(\theta)}$

$= \int \frac{\frac{\sqrt{2}}{3} \sec(\theta) \tan(\theta) d\theta}{\sqrt{2}\tan(\theta)} = \int \frac{1}{3} \sec(\theta) d\theta$

Step 3: Integrate

The integral of $\sec(\theta)$ is:

$\int \sec(\theta) d\theta = \ln \left| \sec(\theta) + \tan(\theta) \right| + C$

So, the solution becomes:

$\frac{1}{3} \ln \left| \sec(\theta) + \tan(\theta) \right| + C$

Step 4: Substitute back the original variable

Recall that:

$u = \frac{\sqrt{2}}{3} \sec(\theta)$

So:

$\sec(\theta) = \frac{3u}{\sqrt{2}} = \frac{3\left(x + \frac{2}{3}\right)}{\sqrt{2}}$

Thus, the integral in terms of $x$ is:

$\frac{1}{3} \ln \left| \frac{3\left(x + \frac{2}{3}\right)}{\sqrt{2}} + \sqrt{\left(\frac{3\left(x + \frac{2}{3}\right)}{\sqrt{2}}\right)^2 - 1} \right| + C$

This is the solution to the integral.

Would you like more details on any specific step or have any further questions?

Here are five related questions to expand your understanding:

1. How can you simplify integrals involving square roots of quadratic expressions?
2. What are the common trigonometric substitutions for different forms of integrals?
3. Can you explain how to convert between different forms of trigonometric identities during integration?
4. How would this integral change if the coefficients in the quadratic were different?
5. How does completing the square help in solving integrals involving square roots?

Tip: Trigonometric substitution is particularly useful for integrals involving square roots of quadratic expressions. Familiarize yourself with the common forms and their corresponding substitutions.