The correct option is C 2(b a)√(x ab x)+c Let ∫ 1 (x b)√( (x a)(b x) ) #160; dx =∫ 1 ( 1)(b x)√({ (b x)+b a}(b x) ) #160; dx =I Substitute ...

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Integration of Irrational Algebraic Fractions - 2

∫dxx-b√x-ab-x...

Question

$\int \frac{d x}{(x - b) \sqrt{(x - a) (b - x)}} =$

- \frac{(b - a)}{2} \sqrt{\frac{b - x}{x - a}} + c

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- (b - a) \sqrt{(b - x) (x - a)} + c

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- \frac{2}{(b - a)} \sqrt{\frac{x - a}{b - x}} + c

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(b - a) \sqrt{(x - b) (x - a)} + c

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Solution

The correct option is C $- \frac{2}{(b - a)} \sqrt{\frac{x - a}{b - x}} + c$
Let $\int \frac{1}{(x - b) \sqrt{(x - a) (b - x)}} d x = \int \frac{1}{(- 1) (b - x) \sqrt{{- (b - x) + b - a} (b - x)}} d x = I$

Substitute

$b - x = \frac{1}{t}$ and $d x = \frac{d t}{t^{2}}$

Therefore
$I = \int \frac{1}{(- 1) (\frac{1}{t}) \sqrt{{\frac{- 1}{t} + b - a} (\frac{1}{t})}} \frac{d t}{t^{2}}$

$= \int \frac{1}{(- 1) \sqrt{{(b - a) t - 1}}} d t$

Now substitute

$(b - a) t - 1 = u$ and $d t = \frac{d u}{(b - a)}$

So integration become

$I = \frac{- 1}{(b - a)} \int \frac{1}{\sqrt{u}} d u$
$= \frac{- 2 \sqrt{u}}{(b - a)} + C$
$= \frac{- 2 \sqrt{(b - a) t - 1}}{(b - a)} + C$
$= \frac{- 2}{(b - a)} \sqrt{\frac{x - a}{b - x}} + C$

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