$\large\frac{dy}{dx}= \frac{x(y^2+4)}{(x^2+1)}$
$\int \large\frac{dy}{y^2+4}=\int \large\frac{x}{x^2+1}$$dx$
$x^2+1=t$
$2xdx=dt$
$\large\frac{1}{2}$$\tan ^{-1}(y/2)=\large\frac{dt}{2t}$$+c$
$\tan ^{-1} \bigg( \large\frac{y}{2}\bigg)$$=\log t+c$
$\tan ^{-1} \bigg( \large\frac{y}{2}\bigg)$$=\log (x^2+1)+c$
Hence a is the correct answer.