if y(x−y)^2 =x, then ∫(dx/(x−3y))=?


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Question Number 175937 by qaz last updated on 09/Sep/22

$i f y {(x - y)}^{2} = x, t h e n \int \frac{d x}{x - 3 y} = ?$
	Answered by Frix last updated on 11/Sep/22

	$y {(x - y)}^{2} = x$ $x^{2} y - 2 {x y}^{2} - x + y^{3} = 0$ $\Rightarrow$ $1 .$ $(2 x y - 2 y^{2} - 1) d x + (x^{2} - 4 x y + 3 y^{2}) d y = 0$ $d x = - \frac{(x - 3 y) (x - y)}{2 x y - 2 y^{2} - 1} d y$ $2 . x = \frac{2 y^{2} + 1 \pm \sqrt{4 y^{2} + 1}}{2 y}$ $\Rightarrow$ $\int \frac{d x}{x - 3 y} = - \int (\frac{1}{2 y} \pm \frac{1}{2 y \sqrt{4 y^{2} + 1}}) d y =$ $= - \frac{\ln ∣ y ∣}{2} \pm \frac{\ln ∣ y ∣ - \ln ∣ 1 - \sqrt{4 y^{2} + 1} ∣}{2} =$ $= - \frac{\ln ∣ 1 \pm \sqrt{4 y^{2} + 1} ∣}{2} + C$
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