Inverse+Integration

Substitue $x = \tan \theta$

$dx = \sec^2\theta d\theta$

This gives,

$\large \int_0^\frac{\pi}{3} \frac{1}{1+tan^2\theta}\sin^{-1}\bigl(\frac{2\tan\theta}{1+\tan^2\theta}\bigr)\times \sec^2\theta d\theta$

$\large = \int_0^\frac{\pi}{3} \sin^{-1}(\sin 2\theta) d\theta$

Now for $0 \leq 2\theta \leq \frac{\pi}{2}; \sin^{-1}(\sin2\theta) = 2\theta$

And for $\frac{\pi}{2} \leq 2\theta \leq \frac{2\pi}{3}; \sin^{-1}(\sin2\theta) = \pi - 2\theta$

This gives,

$\large = \int_0^\frac{\pi}{4} 2\theta d\theta\ + \int_\frac{\pi}{4}^\frac{\pi}{3} (\pi - 2\theta) d\theta$

$\large = \frac{7\pi^2}{72}$

Hence, $a+b+c = 7+2+72 = 81$

$\begin{aligned} I & = \int_0^{\sqrt 3} \frac 1{1+x^2} \sin^{-1}\left(\frac {2x}{1+x^2}\right) dx & \small \color{#3D99F6} \text{Let }x = \tan \frac \theta 2 \implies dx = \frac 12 \sec^2 \frac \theta 2\ d\theta \\ & = \int_0^{\frac {2\pi}3} \frac 1{1 + \tan^2 \frac \theta 2} \sin^{-1} \left(\frac {2\tan \frac \theta 2}{1+\tan^2 \frac \theta 2}\right) \cdot \frac 12 \sec^2 \frac \theta 2\ d\theta \\ & = \frac 12 \int_0^{\frac {2\pi}3} \frac 1{\sec^2 \frac \theta 2} \sin^{-1} \left(\sin \theta \right) \cdot \sec^2 \frac \theta 2\ d\theta \\ & = \frac 12 \int_0^{\frac {2\pi}3} \sin^{-1} (\sin \theta) \ d\theta & \small \color{#3D99F6} \text{Note that } - \frac \pi 2 \le \sin^{-1} x \le \frac \pi 2 \\ & = \frac 12 \left(\int_0^\frac \pi 2 \sin^{-1} (\sin \theta) \ d\theta + \int_\frac \pi 2^{\frac {2\pi}3} \sin^{-1} (\sin (\pi - \theta)) \ d\theta\right) \\ & = \frac 12 \left(\int_0^\frac \pi 2 \theta \ d\theta + \int_\frac \pi 2^{\frac {2\pi}3} (\pi - \theta) \ d\theta\right) \\ & = \frac {\theta^2}4 \ \bigg|_0^\frac \pi 2 + \frac {\pi \theta}2 \ \bigg|_\frac \pi 2^{\frac {2\pi}3} - \frac {\theta^2}4 \ \bigg|_\frac \pi 2^{\frac {2\pi}3} \\ & = \frac {\pi^2}{16} + \frac {\pi^2}3 - \frac {\pi^2}4 - \frac {\pi^2}9 + \frac {\pi^2}{16} \\ & = \frac {7\pi^2}{72} \end{aligned}$

Therefore, $a+b+c = 7+2+72 = \boxed{81}$ .