Question

The value of $\int_0^1 \tan^{-1}\left(\frac{2x - 1}{1 + x - x^2}\right)\,dx$ is

• A. $2$
• B. $-1$
• C. $1$
• D. $0$

Hint:

Whenever you see a definite integral from $0$ to $1$ featuring a symmetric internal structure under the transformation $x \rightarrow 1-x$, look out for complete cancellation! Splitting the terms using arctangent properties almost always creates an odd-symmetric pair that sums to zero under King's property.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem asks for the evaluation of a definite integral involving an inverse trigonometric function from $x = 0$ to $x = 1$.

Step 2: Key Formula or Approach:
1. Use the inverse trigonometric identity to split the fraction inside the argument: $$\tan^{-1}\left(\frac{A - B}{1 + AB}\right) = \tan^{-1}A - \tan^{-1}B$$ 2. Apply the definitive integral property (King's Property): $$\int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx$$

Step 3: Detailed Explanation:
Let's first rewrite the algebraic expression inside the arctangent function. Rearrange the denominator: $$1 + x - x^2 = 1 + x(1 - x)$$ Now, manipulate the numerator to match the factors of the denominator: $$2x - 1 = x - (1 - x)$$ Substituting these representations back, the integral becomes: $$I = \int_0^1 \tan^{-1}\left(\frac{x - (1 - x)}{1 + x(1 - x)}\right) \, dx$$ Using our identity, split this into two separate inverse tangent components: $$I = \int_0^1 \left[ \tan^{-1}(x) - \tan^{-1}(1 - x) \right] \, dx \quad \text{--- (Equation 1)}$$ Now, let's apply King's Property by substituting $x \rightarrow (0 + 1 - x) = 1 - x$: $$I = \int_0^1 \left[ \tan^{-1}(1 - x) - \tan^{-1}(1 - (1 - x)) \right] \, dx$$ $$I = \int_0^1 \left[ \tan^{-1}(1 - x) - \tan^{-1}(x) \right] \, dx \quad \text{--- (Equation 2)}$$ Let's add Equation 1 and Equation 2 together: $$2I = \int_0^1 \left[ \tan^{-1}(x) - \tan^{-1}(1 - x) + \tan^{-1}(1 - x) - \tan^{-1}(x) \right] \, dx$$ Notice that every single term inside the integrand cancels out completely: $$2I = \int_0^1 0 \, dx \implies 2I = 0 \implies I = 0$$

Step 4: Final Answer:
The value of the definite integral is $0$, which corresponds to option (D).