∫(dx/(x(√(1−2x))))


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Question Number 187317 by ajfour last updated on 15/Feb/23

$\int \frac{d x}{x \sqrt{1 - 2 x}}$
	Answered by MJS_new last updated on 15/Feb/23

	$\int \frac{d x}{x \sqrt{1 - 2 x}} =$ $[t = \sqrt{1 - 2 x} \to d x = - \sqrt{1 - 2 x} d t]$ $= 2 \int \frac{d t}{t^{2} - 1} = \ln \frac{t - 1}{t + 1} =$ $= \ln ∣ \frac{1 - \sqrt{1 - 2 x}}{1 + \sqrt{1 - 2 x}} ∣ + C$
	Commented by ajfour last updated on 16/Feb/23

	$t h a n k y o u, i s h a l l m a k e g o o d u s e$ $o f i t!$
	Answered by Humble last updated on 16/Feb/23

	$l e t u = 1 - 2 x$ $\int \frac{1}{\sqrt{u} (u - 1)} d u$ $l e t s = \sqrt{u} ==> u = s^{2}$ $d s = \frac{1}{2 \sqrt{u}} d u ==> d u = 2 \sqrt{u} d s$ $\int \frac{2 \sqrt{u}}{\sqrt{u} (s^{2} - 1)} d s$ $\int \frac{2}{(s^{2} - 1)} d s$ $2 \int \frac{1}{(s^{2} - 1)} d s ==> 2 [- (\frac{l n ∣ s + 1 ∣}{2} - \frac{l n ∣ s - 1 ∣}{2})]$ $s = \sqrt{u} ==> u = \sqrt{1 - 2 x}$ $- l n ∣ \sqrt{1 - 2 x} + 1 ∣ + l n ∣ \sqrt{1 - 2 x} - 1 ∣$ $l n ∣ \sqrt{1 - 2 x} - 1 ∣ - l n ∣ \sqrt{1 - 2 x} + 1 ∣ + C$
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