Question:

Step 1: Set up the integral and identify the function in the numerator. Let us denote: $$I = \int_{-1/\sqrt{3}}^{1/\sqrt{3}} \frac{\cos^{-1}\!\left(\frac{2x}{1+x^2}\right) + \tan^{-1}\!\left(\frac{2x}{1-x^2}\right)}{e^x+1}\,dx$$ Define the numerator function: $$f(x) = \cos^{-1}\!\left(\frac{2x}{1+x^2}\right) + \tan^{-1}\!\left(\frac{2x}{1-x^2}\right)$$ Step 2: Verify that $f(x)$ is an even function. We check that $f(-x) = f(x)$. Since both inverse trigonometric terms in $f(x)$ are even functions (the arguments contain $x$ in even powers or cancel out the sign), we confirm that $f(x)$ is even. Step 3: Apply King's property to simplify the integral. Using the King's property, which states that for an even function $f(x)$ in the denominator with $e^x + 1$: $$I = \int_{-a}^{a} \frac{f(x)}{e^x+1}\,dx = \int_{-a}^{a} \frac{f(x)}{e^{-x}+1}\,dx$$ Adding these two expressions: $$2I = \int_{-1/\sqrt{3}}^{1/\sqrt{3}} f(x) \cdot \left(\frac{1}{e^x+1} + \frac{1}{e^{-x}+1}\right)\,dx$$ Since $\frac{1}{e^x+1} + \frac{1}{e^{-x}+1} = 1$, we have: $$2I = \int_{-1/\sqrt{3}}^{1/\sqrt{3}} f(x)\,dx$$ Step 4: Use the even property to reduce the integration limits. Since $f(x)$ is even: $$2I = 2\int_{0}^{1/\sqrt{3}} f(x)\,dx$$ Therefore: $$I = \int_{0}^{1/\sqrt{3}} f(x)\,dx$$ Step 5: Simplify $\cos^{-1}\!\left(\frac{2x}{1+x^2}\right)$ using substitution. Let $x = \tan\theta$ where $\theta \in (0, \pi/6)$ for $x \in (0, 1/\sqrt{3})$. Then: $$\frac{2x}{1+x^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \frac{2\tan\theta}{\sec^2\theta} = 2\sin\theta\cos\theta = \sin 2\theta$$ Therefore: $$\cos^{-1}\!\left(\frac{2x}{1+x^2}\right) = \cos^{-1}(\sin 2\theta) = \frac{\pi}{2} - 2\theta = \frac{\pi}{2} - 2\tan^{-1}(x)$$ Step 6: Simplify $\tan^{-1}\!\left(\frac{2x}{1-x^2}\right)$ using the double angle formula. For $|x| < 1$, the double angle formula for inverse tangent gives: $$\tan^{-1}\!\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}(x)$$ Step 7: Combine the simplified terms to find $f(x)$. $$f(x) = \cos^{-1}\!\left(\frac{2x}{1+x^2}\right) + \tan^{-1}\!\left(\frac{2x}{1-x^2}\right)$$ $$f(x) = \left(\frac{\pi}{2} - 2\tan^{-1}(x)\right) + 2\tan^{-1}(x) = \frac{\pi}{2}$$ Step 8: Evaluate the integral with the constant function. $$I = \int_{0}^{1/\sqrt{3}} \frac{\pi}{2}\,dx = \frac{\pi}{2} \cdot \left[\,x\,\right]_{0}^{1/\sqrt{3}} = \frac{\pi}{2} \cdot \frac{1}{\sqrt{3}} = \frac{\pi}{2\sqrt{3}}$$ **Final Answer:** The value of the definite integral is $\boxed{\frac{\pi}{2\sqrt{3}}}$, which corresponds to **Option 1**. <div class="key-concept"><strong>Key Concept:</strong> Properties of definite integrals (King's property) and inverse trigonometric identities</div> <div class="trap-box"><strong>Trap:</strong> Forgetting to use the King's property to handle the $e^x+1$ denominator, or incorrectly simplifying the inverse trig expressions.</div>