∫5−xx−2dx=∫5−x−x2+7x−10dx −x2+7x−105−x=A.ddx(−x2+7x−10)+B =−[x2−7x+10]⇒5−x=A(−2x+7)+B =−[(x2)−2(x)(72)+494+10−494]Comparing like coefficients, we get =−[x−722−94]−1=−2A⇒A=12 =(32)2−(x−72)25=7A+B⇒5=72+B⇒B=5−72⇒B=32∴12∫−2x+7−x2+7x−10dx+32∫1−x2+7x−10dx=12[2−x2+7x−10]+32∫1322−x−722dx=−x2+7x−10+32sin−12x−73+c

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$\int \sqrt{[\frac{5 - x}{x - 2}]} d x =$

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$\sqrt{7 x - x^{2} - 10} + \frac{3}{2} \sin^{- 1} (\frac{2 x - 7}{3}) + c$

$\sqrt{7 x - x^{2} - 10} - \frac{3}{2} \sin^{- 1} (\frac{2 x + 7}{3}) + c$

$\sqrt{7 x - x^{2} + 10} + \frac{3}{2} \sin^{- 1} (\frac{2 x - 7}{3}) + c$

$\sqrt{7 x - x^{2} + 10} - \frac{3}{2} \sin^{- 1} (\frac{2 x - 7}{3}) + c$

answer is D.

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Detailed Solution

$\begin{array}{l} \int \sqrt{\frac{5 - x}{x - 2}} d x = \int \frac{5 - x}{\sqrt{- x^{2} + 7 x - 10}} d x - x^{2} + 7 x - 10 \\ 5 - x = A . \frac{d}{d x} (- x^{2} + 7 x - 10) + B = - [x^{2} - 7 x + 10] \\ \Rightarrow 5 - x = A (- 2 x + 7) + B = - [(x^{2}) - 2 (x) (\frac{7}{2}) + \frac{49}{4} + 10 - \frac{49}{4}] \end{array}$ Comparing like coefficients, we get $= - [{(x - \frac{7}{2})}^{2} - \frac{9}{4}]$