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To evaluate the integral:

\[
\int \frac{2x}{1 + 3x^2} \, dx
\]

### Step 1: Simplify the integral
Notice that the numerator \(2x\) is the derivative of the denominator \(1 + 3x^2\), making this a perfect candidate for substitution.

### Step 2: Use substitution
Let:
\[
u = 1 + 3x^2
\]
Then:
\[
\frac{du}{dx} = 6x \quad \Rightarrow \quad du = 6x \, dx
\]
This means:
\[
2x \, dx = \frac{1}{3} \, du
\]

### Step 3: Rewrite the integral
Substitute \(u\) and \(du\) into the integral:
\[
\int \frac{2x}{1 + 3x^2} \, dx = \int \frac{1}{u} \cdot \frac{1}{3} \, du
\]
\[
= \frac{1}{3} \int \frac{1}{u} \, du
\]

### Step 4: Integrate
The integral of \(\frac{1}{u}\) is \(\ln|u|\):
\[
\frac{1}{3} \int \frac{1}{u} \, du = \frac{1}{3} \ln|u| + C
\]

### Step 5: Back-substitute \(u\)
Recall that \(u = 1 + 3x^2\). Substituting back:
\[
\frac{1}{3} \ln|u| + C = \frac{1}{3} \ln|1 + 3x^2| + C
\]

### Final Answer:
\[
\int \frac{2x}{1 + 3x^2} \, dx = \boxed{\frac{1}{3} \ln|1 + 3x^2| + C}
\]

Would you like additional details or have any follow-up questions?

Here are some related questions you might find interesting:
1. How do we handle similar integrals with more complex numerators?
2. What happens if the numerator isn’t a derivative of the denominator?
3. How can substitution be identified quickly in integrals like this?
4. Can this problem be solved using a different integration method?
5. What is the geometric interpretation of this integral?

**Tip:** Always check if the numerator is related to the derivative of the denominator to simplify the integral.

Evaluate the integral \( \int \frac{2x}{1 + 3x^2} \, dx \).

Understand how to evaluate the integral \( \int \frac{2x}{1 + 3x^2} \, dx \) step-by-step using the substitution method. This solution covers substitution techniques, logarithmic integration, and simplifications for students in advanced high school or undergraduate calculus.

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