Let x=\sqrt3\tan\theta\implies dx=\sqrt3\sec^2\theta\ d\theta
\int \frac{dx}{x\sqrt{3+x^2}}
=\int \frac{\sqrt3\sec^2\theta\ d\theta}{\sqrt3\tan\theta\sqrt{3+3\tan^2\theta}}
=\int \frac{\sec^2\theta\ d\theta}{\sqrt3\tan\theta\sqrt{1+\tan^2\theta}}
=\int \frac{\sec^2\theta\ d\theta}{\sqrt3\tan\theta\sec\theta}
=1/\sqrt3\int \frac{\sec\theta\ d\theta}{\tan\theta}
=1/\sqrt3\int \frac{\sec\theta\cos\theta\ d\theta}{\sin\theta}
=1/\sqrt3\int \frac{ d\theta}{\sin\theta}
=1/\sqrt3\int \cosec \thetad\theta
=1/\sqrt3\ln|\tan(\theta/2)|+C
=1/\sqrt3\ln|\tan(1/2\tan^{-1}(x/\sqrt3))|+C