Question:

\textbf{Step 1: Simplify the inverse trigonometric expressions.} Let $x = \tan\theta$. For $x \in \left[-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right]$, we have $\theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$. For the first term: $$ \cos^{-1}\left(\frac{2x}{1+x^2}\right) = \cos^{-1}(\sin(2\theta)) $$ Since $\theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$, we have $2\theta \in \left[-\frac{\pi}{3}, \frac{\pi}{3}\right]$. We use the identity $\sin(y) = \cos\left(\frac{\pi}{2} - y\right)$. $$ \cos^{-1}(\sin(2\theta)) = \cos^{-1}\left(\cos\left(\frac{\pi}{2} - 2\theta\right)\right) $$ Since $2\theta \in \left[-\frac{\pi}{3}, \frac{\pi}{3}\right]$, it follows that $\frac{\pi}{2} - 2\theta \in \left[\frac{\pi}{2} - \frac{\pi}{3}, \frac{\pi}{2} + \frac{\pi}{3}\right] = \left[\frac{\pi}{6}, \frac{5\pi}{6}\right]$. In this interval, $\cos^{-1}(\cos(y)) = y$. Thus, $$ \cos^{-1}\left(\frac{2x}{1+x^2}\right) = \frac{\pi}{2} - 2\theta = \frac{\pi}{2} - 2\tan^{-1}(x) $$ For the second term: $$ \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \tan^{-1}(\tan(2\theta)) $$ Since $2\theta \in \left[-\frac{\pi}{3}, \frac{\pi}{3}\right]$, which is within the principal value range $(-\frac{\pi}{2}, \frac{\pi}{2})$ for $\tan^{-1}$, we have: $$ \tan^{-1}(\tan(2\theta)) = 2\theta = 2\tan^{-1}(x) $$ Summing the terms in the numerator: $$ \left(\frac{\pi}{2} - 2\tan^{-1}(x)\right) + \left(2\tan^{-1}(x)\right) = \frac{\pi}{2} $$ \textbf{Step 2: Rewrite the integral.} The integral becomes: $$ I = \int_{-1/\sqrt{3}}^{1/\sqrt{3}} \frac{\pi/2}{e^x+1}\, dx $$ \textbf{Step 3: Apply the symmetry property.} Let $f(x) = \frac{\pi}{2}$. This is an even function. The integral is of the form $\int_{-a}^{a} \frac{f(x)}{e^x+1}\, dx$. We use the property: $$ \int_{-a}^{a} \frac{g(x)}{e^x+1}\, dx = \int_0^a g(x)\, dx \quad \text{if } g(x) \text{ is an even function.} $$ Proof of the property: Let $I = \int_{-a}^{a} \frac{g(x)}{e^x+1}\, dx$. Using the substitution $x = -t$, $dx = -dt$: $$ I = \int_{a}^{-a} \frac{g(-t)}{e^{-t}+1}\, (-dt) = \int_{-a}^{a} \frac{g(-t)}{e^{-t}+1}\, dt = \int_{-a}^{a} \frac{g(-x)e^x}{1+e^x}\, dx $$ Adding the two expressions for $I$: $$ 2I = \int_{-a}^{a} \left( \frac{g(x)}{e^x+1} + \frac{g(-x)e^x}{e^x+1} \right)\, dx = \int_{-a}^{a} \frac{g(x) + g(-x)e^x}{e^x+1}\, dx $$ Since $g(x)$ is an even function, $g(-x) = g(x)$. $$ 2I = \int_{-a}^{a} \frac{g(x) + g(x)e^x}{e^x+1}\, dx = \int_{-a}^{a} \frac{g(x)(1+e^x)}{e^x+1}\, dx = \int_{-a}^{a} g(x)\, dx $$ Since $g(x)$ is even, $\int_{-a}^{a} g(x)\, dx = 2 \int_0^a g(x)\, dx$. Thus, $2I = 2 \int_0^a g(x)\, dx$, which implies $I = \int_0^a g(x)\, dx$. \textbf{Step 4: Evaluate the integral.} Applying the property with $a = \frac{1}{\sqrt{3}}$ and $f(x) = \frac{\pi}{2}$: $$ I = \int_0^{1/\sqrt{3}} \frac{\pi}{2}\, dx $$ $$ I = \left[ \frac{\pi}{2} x \right]_0^{1/\sqrt{3}} $$ $$ I = \frac{\pi}{2} \left( \frac{1}{\sqrt{3}} - 0 \right) = \frac{\pi}{2\sqrt{3}} $$ Numerically, $$ I \approx \frac{3.1415926535}{2 \times 1.732050810} \approx \frac{3.1415926535}{3.464101620} \approx 0.90689968 $$ Rounding to four decimal places, the value is $0.9069$.