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Question #41184

Difficulty: 8/10 | 🟢 Free

The Question

Evaluate the integral: $I = \int_{0}^{\sqrt{3}} \left( \arctan\left(\frac{2x}{1-x^2}\right) + \frac{2x}{1+x^2} \right) dx$

Options

A.
$\pi \left( 1 - \frac{1}{\sqrt{3}} \right)$
B.
$-\frac{\pi}{\sqrt{3}}$
C.
$\frac{2\pi}{\sqrt{3}}$
D.
$\pi \left( 1 + \frac{1}{\sqrt{3}} \right)$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Inverse Function Periodicity|Factor-Product Conversion|Substitution (Change of Variable)

Why This Question Is Tricky

Common Mistakes

This problem tests the student's mastery of inverse trigonometric identities and their principal value branches.

The core trap lies in the factor-product conversion $\arctan(\frac{2x}{1-x^2}) = 2\arctan x$ , which is only valid for $|x| < 1$ .

Students who blindly apply this identity across the entire interval $[0, \sqrt{3}]$ will incorrectly obtain $\frac{2\pi}{\sqrt{3}}$ .

Alternatively, students who recognize the integrand as the derivative of $x \arctan(\frac{2x}{1-x^2})$ might blindly apply the Fundamental Theorem of Calculus without splitting the integral at the jump discontinuity $x=1$ , leading to the trap answer $-\frac{\pi}{\sqrt{3}}$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	Inverse Function Periodicity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free