Question

\(\int \frac{dx}{x^2 + 4} \)

• A. \( \frac{1}{4} \tan^{-1} \frac{x}{4} + k \)
• B. \( \frac{1}{2} \tan^{-1} \frac{x}{2} + k \)
• C. \( \frac{1}{2} \tan^{-1} \frac{2x}{x} + k \)
• D. \( 2 \tan^{-1} \frac{x}{2} + k \)

Hint:

Remember that the derivative of \( \tan^{-1} x \) is \( \frac{1}{1+x^2} \), and when integrating, use the corresponding standard form.

Answer 1

The Correct Option is B

We are given the integral: \[ I = \int \frac{dx}{x^2 + 4} \] This is a standard integral of the form \( \int \frac{dx}{x^2 + a^2} \), which can be solved using the formula: \[ \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \frac{x}{a} + C \] In this case, \( a = 2 \), so the solution becomes: \[ I = \frac{1}{2} \tan^{-1} \frac{x}{2} + C \] Thus, the answer is \( \frac{1}{2} \tan^{-1} \frac{x}{2} + k \).