a-x-1-ax-dx-a-gt-0-

Question Number 127190 by bramlexs22 last updated on 27/Dec/20

\int \frac{\sqrt{a} - \sqrt{x}}{1 - \sqrt{a x}} d x = ?; a > 0

Answered by Dwaipayan Shikari last updated on 27/Dec/20

\Rightarrow a x = u^{2} \Rightarrow a = 2 u \frac{d u}{d x}

2 \int \frac{\sqrt{a} - \frac{u}{\sqrt{a}}}{1 - u} . \frac{u}{a} d u = \frac{2}{\sqrt{a^{3}}} \int \frac{a - u}{1 - u} d u = \frac{2}{\sqrt{a^{3}}} u + \frac{2}{\sqrt{a^{3}}} \int \frac{a - 1}{1 - u} d u

= \frac{2}{\sqrt{a^{3}}} u - \frac{2 (a - 1)}{\sqrt{a^{3}}} l o g (1 - u) + C = \frac{2 \sqrt{x}}{a} - \frac{2 (a - 1)}{\sqrt{a^{3}}} l o g (1 - \sqrt{a x}) + C

Answered by liberty last updated on 27/Dec/20

I = \int \frac{\sqrt{a} - \sqrt{x}}{1 - \sqrt{a x}} d x; t a k e w = 1 - \sqrt{a x}

d x = - \frac{2 (1 - w)}{a} d w

I = \int \frac{\sqrt{a} - \frac{1 - w}{\sqrt{a}}}{w} (- \frac{2 (1 - w)}{a}) d w

I = - \frac{2}{a \sqrt{a}} \int (\frac{a - 1}{w} + (2 - a) - w) d w

I = - \frac{2}{a \sqrt{a}} [(a - 1) \ln w + (2 - a) w - \frac{w^{2}}{2}] + c

I = \frac{1}{a \sqrt{a}} [2 (1 - a) \sqrt{a x} + a x - 2 (a - 1) \ln (1 - \sqrt{a x})] + c

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