Tricky Definite Integral

Relevant wiki: Half Angle Tangent Substitution

Similar solution with @Benny Joseph 's

$\begin{aligned} I & = \int_0^{\sqrt 3} \frac 1{1+x^2} \sin^{-1} \left(\frac {2x}{1+x^2}\right) dx & \small \color{#3D99F6} \text{Let }x = \tan \theta \implies dx = \sec^2 \theta \ d\theta \\ & = \int_0^\frac \pi 3 \frac {\sec^2 \theta}{1+\tan^2 \theta} \sin^{-1} \left(\frac {2\tan \theta}{1+\tan^2 \theta}\right) d\theta & \small \color{#3D99F6} \text{Half angle tangent substitution: }\sin u = \dfrac {2\tan \frac u2}{1+ \tan^2 \frac u2} \\ & = \int_0^\frac \pi 3 \frac {\sec^2 \theta}{\sec^2 \theta} \sin^{-1} \left(\sin {\color{#3D99F6}2 \theta} \right) \ d\theta & \small \color{#3D99F6} \text{Let }\phi = 2 \theta \implies d \phi = 2 \ d\theta \\ & = \frac 12 \int_0^{\color{#D61F06}\frac {2\pi} 3} \sin^{-1} \left(\sin {\color{#3D99F6}\phi} \right) \ d\phi & \small \color{#D61F06} \text{Note that }\sin^{-1} u \text{ is defined } \in \left[-\frac \pi 2, \frac \pi 2\right] \\ & = \frac 12 \left(\int_{\color{#D61F06}0} ^{\color{#D61F06}\frac \pi 2} \sin^{-1} (\sin \phi) \ d\phi + \int_{\color{#D61F06}\frac \pi 3}^{\color{#D61F06}\frac \pi 2} \sin^{-1} (\sin \phi) \ d\phi \right) \\ & = \frac 12 \left(\int_0^\frac \pi 2 \phi \ d\phi + \int_\frac \pi 3^\frac \pi 2 \phi \ d\phi \right) \\ & = \frac 12 \left(\frac {\phi^2} 2 \bigg|_0^\frac \pi 2 + \frac {\phi^2} 2 \bigg|_\frac \pi 3^\frac \pi 2 \right) \\ & = \frac 14 \left(\frac {\pi^2} 4 + \frac {\pi^2} 4 - \frac {\pi^2} 9 \right) \\ & = \boxed{\dfrac {7\pi^2}{72}} \end{aligned}$

put $\tan ^{-1}x\ =\theta$ . Then the integral becomes $\int _0^{\frac{\pi }{3}}\sin ^{-1}\left(\sin 2\theta \right)d\theta$ . If we notice carefully, we find $\sin ^{-1}\left(\sin 2\theta \right)$ changes as x varies from zero to pi/3. Hence the step $\int _0^{\frac{\pi }{4}}\sin ^{-1}\left(\sin 2\theta \right)d\theta +\int _{\frac{\pi }{4}}^{\frac{\pi }{3}}\sin ^{-1}\left(\sin 2\theta \right)d\theta$ which is $\int _0^{\frac{\pi }{4}}2\theta d\theta +\int _{\frac{\pi }{4}}^{\frac{\pi }{3}}\left(\pi -2\theta \right)d\theta$