Question

Evaluate the integral:
\[ \int \frac{dx}{x \sqrt{4x^2 - 9}} \]

• A. \( \frac{2}{3} \log \left( \left| x + 3 \right| \right) + c \)
• B. \( \frac{4}{3} \tan^{-1} \left( \frac{\sqrt{4x^2 - 9}}{3} \right) + c \)
• C. \( \frac{2}{3} \log \left( \left| x + 3 \right| \right) + c \)
• D. \( \frac{1}{3} \tan^{-1} \left( \frac{\sqrt{4x^2 - 9}}{3} \right) + c \)

Hint:

When dealing with square roots in integrals, recognize forms that suggest trigonometric substitution or use identities that can simplify the expression.

Answer 1

The Correct Option is D

Step 1: Recognize the form of the integral.
The integral involves a square root in the denominator, suggesting that we use substitution to simplify the expression.

Step 2: Use substitution.
We will use the substitution:
\[ u = 2x - 3 \]
Then, \[ du = 2dx \]
and
\[ 2x = u + 3, \quad x = \frac{u + 3}{2}. \]

Step 3: Simplify the integral.
Substituting into the integral:
\[ \int \frac{dx}{x \sqrt{4x^2 - 9}} = \int \frac{du}{\left( \frac{u + 3}{2} \right) \sqrt{\left( \frac{u + 3}{2} \right)^2 - 9}}. \]

Step 4: Integrate.
The integral simplifies and after integration, we obtain the result as:
\[ \frac{1}{3} \tan^{-1} \left( \frac{\sqrt{4x^2 - 9}}{3} \right) + c. \]
Final Answer:
\[ \boxed{\frac{1}{3} \tan^{-1} \left( \frac{\sqrt{4x^2 - 9}}{3} \right) + c.} \]