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Find the solution : $ (1+x^4) \large\frac{dy}{dx}$$-4x^3y =(x^5+x) \tan ^{-1} (x^2) \qquad y(1)=\pi$

$(a)\;y=\frac{1+x^4)}{4} \tan ^{-1} (x^2)^2+c \\ (b)\;y=\frac{1+x^4)}{4} \tan ^{-1} x^2+c \\ (c)\;y=x^4 \tan ^{-1} (x^2)^2+c\\ (d)\;y=(1+x^4)^2+c $

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