The integral to solve is:

$\int \frac{1}{\sqrt{7x^2 - 1}} \, dx$

Step-by-Step Solution:

To solve this integral, we will use a trigonometric substitution. Notice that the expression inside the square root is in the form of $a x^2 - b$, which suggests a substitution involving a hyperbolic function or a trigonometric function.

Step 1: Substitution

Consider the substitution:

$x = \frac{1}{\sqrt{7}} \sec(\theta)$

This implies:

$dx = \frac{1}{\sqrt{7}} \sec(\theta) \tan(\theta) \, d\theta$

Substitute into the integral:

$7x^2 - 1 = 7 \left( \frac{1}{\sqrt{7}} \sec(\theta) \right)^2 - 1 = 7 \frac{1}{7} \sec^2(\theta) - 1 = \sec^2(\theta) - 1 = \tan^2(\theta)$

Thus:

$\sqrt{7x^2 - 1} = \sqrt{\tan^2(\theta)} = \tan(\theta)$

Step 2: Rewrite the Integral

Substitute all of this into the integral:

$\int \frac{1}{\sqrt{7x^2 - 1}} \, dx = \int \frac{1}{\tan(\theta)} \cdot \frac{1}{\sqrt{7}} \sec(\theta) \tan(\theta) \, d\theta$

This simplifies to:

$\frac{1}{\sqrt{7}} \int \sec(\theta) \, d\theta$

Step 3: Integrate

The integral of $\sec(\theta)$ is:

$\int \sec(\theta) \, d\theta = \ln | \sec(\theta) + \tan(\theta) | + C$

Thus:

$\frac{1}{\sqrt{7}} \ln | \sec(\theta) + \tan(\theta) | + C$

Step 4: Substitute Back

Recall the original substitution $x = \frac{1}{\sqrt{7}} \sec(\theta)$, therefore:

$\sec(\theta) = \sqrt{7} x, \quad \tan(\theta) = \sqrt{7x^2 - 1}$

Substitute back:

$\frac{1}{\sqrt{7}} \ln \left| \sqrt{7} x + \sqrt{7x^2 - 1} \right| + C$

Final Answer:

$\int \frac{1}{\sqrt{7x^2 - 1}} \, dx = \frac{1}{\sqrt{7}} \ln \left| \sqrt{7} x + \sqrt{7x^2 - 1} \right| + C$

Would you like more details or have any questions?

Related Questions:

1. How do you solve integrals involving expressions like $\sqrt{ax^2 + b}$?
2. What is the integral of $\frac{1}{x^2 - a^2}$?
3. How do you perform hyperbolic substitutions in integrals?
4. Can you explain the integrals of hyperbolic functions like $\cosh(x)$ and $\sinh(x)$?
5. What is the difference between using trigonometric and hyperbolic substitutions for integrals?

Tip:

When integrating functions involving $\sqrt{ax^2 - b}$ or similar forms, trigonometric substitutions are a powerful tool to simplify the integrals!