$\large\frac{dy}{dx} -\frac{4x^3y}{1+x^4} $$=x \tan ^{-1} (x^2)$
I.F= $e^{-\int \large\frac{4x^3}{1+x^4}dx}$
$\qquad= e^{- \log (1+x^4)}$
$\qquad= \large\frac{1}{1+x^4}$
So, $y. \large\frac{1}{1+x^4}=\int \large\frac{x \tan ^{-1} (x^2)dx}{1+x^4}$
$y. \large\frac{1}{1+x^4} =\large\frac{1}{2} \frac{(\tan ^{-1} (x^2))^2}{2}$$+c$
$y= \large\frac{(1+x^4)}{4} $$\tan^{-1}(x^2)^2+c$
Hence a is the correct answer.