Question

Evaluate the definite integral $$\int_{1}^{3} \left[ \tan^{-1}\left(\frac{x}{x^2-1}\right) + \tan^{-1}\left(\frac{x^2-1}{x}\right) \right] dx$$

• A. $\pi$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{2}$
• D. $2\pi$

Hint:

Whenever you see two complicated inverse trigonometric terms added together where the numerator and denominator are flipped, don't try to integrate them individually! They will almost always collapse directly into the constant $\frac{\pi}{2}$ via standard identities.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem asks for the value of a definite integral spanning from 1 to 3, where the integrand contains a sum of two inverse trigonometric arc-tangent functions.

Step 2: Key Formula or Approach:
We can use the reciprocal identity for inverse trigonometric functions:
$$\tan^{-1}\left(\frac{1}{\theta}\right) = \cot^{-1}\theta \quad (\text{for } \theta > 0)$$ Combined with the standard complementary angle identity:
$$\tan^{-1}\theta + \cot^{-1}\theta = \frac{\pi}{2}$$

Step 3: Detailed Explanation:
Let's consider the second term of the integrand, $\tan^{-1}\left(\frac{x^2-1}{x}\right)$. Over the given interval of integration $x \in [1, 3]$, the argument is non-negative. Applying the reciprocal identity yields:
$$\tan^{-1}\left(\frac{x^2-1}{x}\right) = \cot^{-1}\left(\frac{x}{x^2-1}\right)$$ Now substitute this back into the full integrand expression:
$$I = \int_{1}^{3} \left[ \tan^{-1}\left(\frac{x}{x^2-1}\right) + \cot^{-1}\left(\frac{x}{x^2-1}\right) \right] dx$$ Using the identity $\tan^{-1}\theta + \cot^{-1}\theta = \frac{\pi}{2}$, the entire complex bracketed expression simplifies to a simple constant value:
$$I = \int_{1}^{3} \left( \frac{\pi}{2} \right) dx$$ $$I = \frac{\pi}{2} [x]_{1}^{3} = \frac{\pi}{2} (3 - 1)$$ $$I = \frac{\pi}{2} \times 2 = \pi$$ This matches option (A).

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