Question

Evaluate the integral \[ \int_{0}^{1} \tan^{-1}\!\left( \frac{2x - 1}{1 + x - x^2} \right) dx \]

• A. \( 0 \)
• B. \( \dfrac{\pi}{6} \)
• C. \( 1 \)
• D. \( \dfrac{\pi}{4} \)

Hint:

For integrals involving inverse trigonometric functions over symmetric limits, always try the substitution \( x \to 1-x \).

Answer 1

The Correct Option is A

Step 1: Use the property of definite integrals.
Let \[ I = \int_{0}^{1} \tan^{-1}\!\left( \frac{2x - 1}{1 + x - x^2} \right) dx \] Using the property \[ I = \int_{0}^{1} f(x)\,dx = \int_{0}^{1} f(1-x)\,dx \]

Step 2: Replace \( x \) by \( 1-x \).
\[ I = \int_{0}^{1} \tan^{-1}\!\left( \frac{1 - 2x}{1 + x - x^2} \right) dx \]

Step 3: Add both expressions.
\[ 2I = \int_{0}^{1} \left[ \tan^{-1}\!\left( \frac{2x - 1}{1 + x - x^2} \right) + \tan^{-1}\!\left( \frac{1 - 2x}{1 + x - x^2} \right) \right] dx \]

Step 4: Use the identity \( \tan^{-1}a + \tan^{-1}(-a) = 0 \).
\[ 2I = 0 \Rightarrow I = 0 \]

Step 5: Conclusion.
\[ \boxed{0} \]