Find a particular solution satisfying the given condition $(1+x^2)\large\frac{dy}{dx}$$+2xy=\large\frac{1}{1+x^2}$$;y=0\;when\;x=1$

Question

$\begin{array}{1 1}(A)\;y(1+x^2) = \tan^{-1} x -\large\frac{ \pi}{2} \\(B)\;y(1+x^2) = \tan^{-1} x -\large\frac{ \pi}{4} \\(C)\;y(1+x^2) = \tan^{-1} x -\large\frac{ 2\pi}{3} \\ (D)\;y(1+x^2) = \tan^{-1} x +\large\frac{ \pi}{4} \end{array} $