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.