Solve:
$(1 - x^{2}) (1 - y) d x = x y (1 + y) d y$

ln (x) (1 - y)^{2} = c + \frac{1}{2} y^{2} - 2 y + \frac{1}{2} x^{2}

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ln (x) (1 - y)^{2} = c - \frac{1}{2} y^{2} - 2 y + \frac{1}{2} x^{2}

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ln (x) (1 - y)^{2} = c - \frac{1}{2} y^{2} + 2 y + \frac{1}{2} y^{2}

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ln (x) (1 + y)^{2} = c + \frac{1}{2} y^{2} - 2 y + \frac{1}{2} x^{2}

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Solution

The correct option is B $ln (x) (1 - y)^{2} = c - \frac{1}{2} y^{2} - 2 y + \frac{1}{2} x^{2}$
$(1 - x^{2}) (1 - y) d x = x y (1 + y) d y$
$\frac{1 - x^{2}}{x} d x = \frac{y^{2} + y}{1 - y} d y$
Integrating both sides, we get
$log x - \frac{x^{2}}{2} = - \int (y + 2 + \frac{2}{y - 1}) d y$
$\Rightarrow log x - \frac{x^{2}}{2} = - \frac{y^{2}}{2} - 2 y + 2 log (y - 1)$
$log (x) (1 - y)^{2} = c - \frac{1}{2} y^{2} - 2 y + \frac{1}{2} x^{2}$

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Solve:(1−x2)(1−y)dx=xy(1+y)dy

The correct option is B ln(x)(1−y)2=c−12y2−2y+12x2(1−x2)(1−y)dx=xy(1+y)dy1−x2xdx=y2+y1−ydyIntegrating both sides, we getlogx−x22=−∫(y+2+2y−1)dy⇒logx−x22=−y22−2y+2log(y−1)log(x)(1−y)2=c−12y2−2y+12x2

Solve:
$(1 - x^{2}) (1 - y) d x = x y (1 + y) d y$