If $y=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\sec ^{-1}\left(\frac{1+x^2}{1-x^2}\right)$ then the value of $\frac{d y}{d x}$ at $x=\sqrt{3}$ is

$$ \int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{x}{1+\sin x} d x= $$

$$ \int_0^2 \frac{3 x+1}{x^2+4} d x= $$

$$ \int \frac{\sin 2 x \cos 2 x}{\sqrt{9-\cos ^4 2 x}} d x= $$