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Solve : $(x^2+1) dy =(4x+xy^2)dx$

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

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