It looks like you're trying to compute the integral:  

\[
\int \frac{dx}{3x^2 - 5}
\]

### Solution:
We rewrite the integral in standard form:

\[
I = \int \frac{dx}{3x^2 - 5}
\]

#### Step 1: Use substitution  
Let \( u = 3x^2 - 5 \), so that:

\[
\frac{du}{dx} = 6x
\]

Rearranging,

\[
du = 6x \, dx
\]

Since there's no \( x \) in the numerator, we proceed differently.

#### Step 2: Express as standard integral  
We rewrite the denominator to match the standard integral form:

\[
\int \frac{dx}{a x^2 + b}
\]

which has the known solution:

\[
\frac{1}{\sqrt{a}} \tan^{-1} \left(\frac{x}{\sqrt{b/a}}\right) + C
\]

Comparing with our integral:

\[
\int \frac{dx}{3x^2 - 5}
\]

we set:

\[
a = 3, \quad b = -5
\]

This integral resembles the form:

\[
\int \frac{dx}{ax^2 + c}
\]

For cases where \( c \) is negative, we use the logarithm-based solution.

#### Step 3: Solve using standard formula  
Rewriting:

\[
\int \frac{dx}{3(x^2 - \frac{5}{3})}
\]

Using the standard integral:

\[
\int \frac{dx}{a(x^2 - c)} = \frac{1}{a\sqrt{c}} \ln \left| \frac{x - \sqrt{c}}{x + \sqrt{c}} \right| + C
\]

where \( c = \frac{5}{3} \), we get:

\[
I = \frac{1}{\sqrt{15}} \ln \left| \frac{x - \sqrt{5/3}}{x + \sqrt{5/3}} \right| + C
\]

### Final Answer:
\[
\frac{1}{\sqrt{15}} \ln \left| \frac{x - \sqrt{5/3}}{x + \sqrt{5/3}} \right| + C
\]

Would you like a step-by-step breakdown or an alternative method?  

### Related Questions:
1. How do you integrate functions of the form \( \frac{dx}{ax^2 + b} \)?
2. What is the role of substitution in solving rational integrals?
3. Can this integral be solved using partial fractions?
4. How does completing the square help in integration problems?
5. What is the connection between this integral and inverse trigonometric functions?

#### **Tip:**  
When integrating rational functions, always check if the denominator can be factored or rewritten into a standard form to simplify the process.

This page provides a step-by-step solution to the integral \int \frac{dx}{3x^2 - 5}. The process includes substitution, recognizing standard forms, and solving using logarithmic solutions. Suitable for advanced calculus students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation