Solution of $(x y^{2} + x) d x + (y x^{2} + y) d y = 0$

(x^{2} + 1) (y^{2} + 1) = c

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(x y + 1) (x y - 1) = c

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(x^{3} + 1) (y^{3} + 1) = c

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(1 - x^{2}) (1 - y^{2}) = c

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Solution

The correct option is A $(x^{2} + 1) (y^{2} + 1) = c$
$\Rightarrow x (y^{2} + 1) d x + y (x^{2} + 1) d y = 0$
$\Rightarrow (\frac{2 x}{x^{2} + 1}) d x + (\frac{2 y}{y^{2} + 1}) d y = 0$
let $\begin{matrix} t = x^{2} + 1 > & k = y^{2} + 1 \Rightarrow d t = 2 x d x & \Rightarrow d k = 2 y d y \end{matrix}$
$\Rightarrow \int \frac{d t}{t} + \int \frac{d k}{k} = l o g c$
$\Rightarrow k t = c$
$\Rightarrow (x^{2} + 1) (y^{2} + 1) = c$

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