Step 1: Identifying the Function for Integration

We start by identifying the function we aim to integrate, which is \( \frac{1}{(3x + 1)^2} \). This function is a rational function with a quadratic term in the denominator.

Step 2: Setting Up the Integral

We set up the integral with the given limits, which are \( x = 0 \) and \( x = \infty \):

\[ \int_{0}^{\infty} \frac{1}{(3x + 1)^2} \, dx \]

Step 3: Applying Limits for the Improper Integral

Since the integral has an infinite upper limit, it is an improper integral. We use limits to evaluate it:

\[ \lim_{{b \to \infty}} \int_{0}^{b} \frac{1}{(3x + 1)^2} \, dx \]

Step 4: Finding the Antiderivative

The antiderivative of \( \frac{1}{(3x + 1)^2} \) is \( -\frac{1}{3(3x + 1)} + C \), where \( C \) is the constant of integration.

Step 5: Applying the Upper and Lower Limits

We apply the upper and lower limits to the antiderivative:

\[ \left[-\frac{1}{3(3x + 1)}\right]_{0}^{b} \]

Step 6: Evaluating the Limits

After applying the limits, we find:

\[ \lim_{{b \to \infty}} \left(-\frac{1}{3(3b + 1)} + \frac{1}{3(3 \cdot 0 + 1)}\right) = 0 + \frac{1}{3} = \frac{1}{3} \]

Step 7: Final Result and Interpretation

The integral converges to \( \frac{1}{3} \). This result is interesting because it shows that the function has a finite integral over the given range, highlighting its rapid decay as \( x \) approaches infinity.