Differentiation

Let $\tan^{-1}x=\theta$, $\theta \in (-\pi/2,\pi/4) \cup (\pi/4,\pi/2)$ $f(x)=(\sin\theta+\cos\theta)^2-1$ $=\sin 2\theta = \frac{2x}{1+x^2}$ Now, $\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ $=-\frac{1}{1+x^2},|x| > 1$ Since, we can integrate only in the continuous interval. So we have to take integral in two cases separtely namely for x < -1 and for x > 1. $\Rightarrow y=\begin{cases} -\tan^{-1}x+c_1, & x > 1 \\ -\tan^{-1}+c_2, & x < -1 \end{cases}$ So, $c_1=\frac{\pi}{2}$ as $y(\sqrt{3})=\frac{\pi}{6}$ But we cannot find $c_2$ as we do not have any other additional information for x < -1. So, all of the given options may be correct as $c_2$ is unknown so, it should be bonus.