The integral provided in the image is:

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

Solution:

1. Substitution Method: Let $u = 1 + 3x^2$. Then, the derivative of $u$ with respect to $x$ is: $\frac{du}{dx} = 6x \quad \implies \quad du = 6x \, dx \quad \implies \quad x \, dx = \frac{1}{6} \, du.$

   Substituting $u$ into the integral: $\int \frac{2x}{1 + 3x^2} \, dx = \int \frac{2}{u} \cdot \frac{1}{6} \, du = \frac{1}{3} \int \frac{1}{u} \, du.$

2. Integrate: The integral of $\frac{1}{u}$ is $\ln|u|$. Thus: $\frac{1}{3} \int \frac{1}{u} \, du = \frac{1}{3} \ln|u| + C.$

3. Back-substitute $u$: Recall $u = 1 + 3x^2$. Substituting back: $\int \frac{2x}{1 + 3x^2} \, dx = \frac{1}{3} \ln|1 + 3x^2| + C.$

Final Answer:

$\int \frac{2x}{1 + 3x^2} \, dx = \frac{1}{3} \ln(1 + 3x^2) + C.$

Let me know if you'd like further details or have any questions! Here are related questions for you to explore:

1. How does substitution work for more complex integrals?
2. Why is the absolute value used in $\ln|u|$ during integration?
3. What other methods can solve this integral, besides substitution?
4. How do we handle definite integrals for this problem?
5. Can this integral be solved using partial fractions?

Tip: Substitution is a powerful technique—look for expressions where the derivative of part of the denominator or numerator is present elsewhere in the integral.