Rearranging the terms of the differential equation, we get $\frac{dx}{dt}+x\left ( \frac{-t}{1-t^{2}} \right )=\frac{1}{1-t^{2}}$
This is the standard linear Differential equation form for which the integrating factor is: $e^{\int \frac{-t}{1-t^{2}}dt}=e^{\frac{1}{2}\times \ln_{e}\left ( 1-t^{2} \right )}=\sqrt{1-t^{2}}$
So, solution to this D.E. is : $x\sqrt{1-t^{2}}=\int \frac{1}{1-t^{2}}\times \sqrt{1-t^{2}}dt$
$\Rightarrow$ let $t=\sin \theta$ $\Rightarrow$ $dt=\cos\theta d\theta$
$\Rightarrow$ $x\sqrt{1-t^{2}}=\int \frac{\cos\theta }{\cos\theta }d\theta$ $\Rightarrow$ $x\sqrt{1-t^{2}}=\sin^{-1}t+c$
Given that at $t=0, x=1$. So, $1\left ( 1 \right )=\sin^{-1}\left ( 0 \right )+c \Rightarrow c=1$
$\therefore$ Solution to the D.E. is $x\sqrt{1-t^{2}}=1+\sin^{-1}(t)$
For $t=\frac{1}{2}$, $x\left ( \frac{\sqrt{3}}{2} \right )=\sin^{-1}(\frac{1}{2})+1$ $\Rightarrow$ $x=\frac{2}{\sqrt{3}}\left ( \frac{\pi }{6}+1 \right )$
Option A is correct.