The given integral is:

$\int \frac{x}{12x^2 - 4} \, dx$

Solution

Step 1: Simplify the denominator

Factor out $4$ from the denominator: $12x^2 - 4 = 4(3x^2 - 1)$ Thus, the integral becomes: $\int \frac{x}{4(3x^2 - 1)} \, dx = \frac{1}{4} \int \frac{x}{3x^2 - 1} \, dx$

Step 2: Substitution

Let: $u = 3x^2 - 1 \quad \text{so that} \quad du = 6x \, dx \quad \text{or} \quad \frac{1}{6} du = x \, dx$

Substitute $u$ and $du$ into the integral: $\frac{1}{4} \int \frac{x}{3x^2 - 1} \, dx = \frac{1}{4} \int \frac{1}{u} \cdot \frac{1}{6} \, du = \frac{1}{24} \int \frac{1}{u} \, du$

Step 3: Integrate

The integral of $\frac{1}{u}$ is $\ln|u|$: $\frac{1}{24} \int \frac{1}{u} \, du = \frac{1}{24} \ln|u| + C$

Step 4: Back-substitute $u$

Substitute $u = 3x^2 - 1$ back into the solution: $\frac{1}{24} \ln|u| + C = \frac{1}{24} \ln|3x^2 - 1| + C$

Final Answer:

$\int \frac{x}{12x^2 - 4} \, dx = \frac{1}{24} \ln|3x^2 - 1| + C$

Would you like a detailed explanation of any step?

Related Questions:

1. How do you handle integrals with quadratic expressions in the denominator?
2. What is the importance of substitution in integration problems?
3. How can you determine when to use logarithmic functions in integrals?
4. What are other common examples of integrals involving $x / (ax^2 + b)$?
5. How does the constant factor $1/4$ affect the integration process?

Tip:

Always check if the denominator can be factored or simplified before proceeding with substitution in integrals.