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The solution of differential equation $(x^2-1) \large\frac{dy}{dx}$$+2xy =\large\frac{1}{x^2-1}$

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

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