Let $x = \tan\theta$, $dx = \sec^2\theta\,d\theta$, $\sqrt{1+x^2}=\sec\theta$.

$\displaystyle\int\frac{\tan^2\theta}{\sec\theta}\sec^2\theta\,d\theta = \int\sec\theta\tan^2\theta\,d\theta = \int\sec^3\theta - \sec\theta\,d\theta$

Using $\int\sec^3\theta\,d\theta = \tfrac12(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|)$:

Bounds: $x=0\Rightarrow\theta=0$; $x=\sqrt3\Rightarrow\theta=\pi/3$, $\sec(\pi/3)=2$

Result $= \bigl[\tfrac12\sec\theta\tan\theta - \tfrac12\ln|\sec\theta+\tan\theta|\bigr]_0^{\pi/3} = \tfrac12(2)(\sqrt3)-\tfrac12\ln(2+\sqrt3) = \sqrt3 - \ln(2+\sqrt3)$

Answer: A.

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