Question

\( \int_0^1 \frac{dx}{1+x^2} \) is equal to:

• A. \( \frac{\pi}{2} \)
• B. \( \frac{\pi}{4} \)
• C. \( \frac{\pi}{3} \)
• D. \( \frac{\pi}{6} \)

Hint:

Integrals involving \( \frac{1}{1+x^2} \), \( \frac{1}{\sqrt{1-x^2}} \), and \( \frac{1}{x\sqrt{x^2-1}} \) are high-frequency exam questions. Always double-check your inverse trig values.

Answer 1

The Correct Option is B

Step 1: Key Formula or Approach:
The integral \( \int \frac{1}{1+x^2} dx \) is a standard integral resulting in \( \tan^{-1} x \).
Step 2: Detailed Explanation:
Using the fundamental theorem of calculus:
\[ \int_0^1 \frac{1}{1+x^2} dx = [\tan^{-1} x]_0^1 \] Evaluate at the upper limit: \( \tan^{-1}(1) = \frac{\pi}{4} \).
Evaluate at the lower limit: \( \tan^{-1}(0) = 0 \).
Subtracting the values:
\[ \frac{\pi}{4} - 0 = \frac{\pi}{4} \] Step 3: Final Answer:
The integral evaluates to \( \frac{\pi}{4} \).