Question

\[ \int \frac{1}{(x^2 + 1)^{\frac{3}{4}}} \, dx \] is equal to

• A. \( \sec^{-1} \left( \frac{x^2 + 1}{\sqrt{2x}} \right) + c \)
• B. \( \frac{1}{\sqrt{2x}} \sec^{-1} \left( \frac{1}{\sqrt{2}} \right) + c \)
• C. \( \frac{1}{\sqrt{2x}} \sec^{-1} \left( \frac{1}{\sqrt{2}} \right) + c \)
• D. None of these

Hint:

For integrals involving powers of \( x^2 + 1 \), use trigonometric substitution to simplify the expression.

Answer 1

The Correct Option is B

Step 1: Use substitution for integration.
We use appropriate trigonometric substitution to solve the integral.
Step 2: Conclusion.
The integral evaluates to \( \sec^{-1} \left( \frac{x^2 + 1}{\sqrt{2x}} \right) + c \). Final Answer: \[ \boxed{\sec^{-1} \left( \frac{x^2 + 1}{\sqrt{2x}} \right) + c} \]