If I=∫5−x2+xdx, then I equals

Put 2+x=t2, so that dx=2t dt, thenI=∫7−t2t(2t)dt =2∫7−t2dt =t7−t2+7sin−1(t7)+C=x+25−x+7sin−1(x+27)+C

If I=∫5−x2+xdx, then I equals

Put 2+x=t2, so that dx=2t dt, thenI=∫7−t2t(2t)dt =2∫7−t2dt =t7−t2+7sin−1(t7)+C=x+25−x+7sin−1(x+27)+C

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If $I = \int \sqrt{\frac{5 - x}{2 + x}} d x,$ then $I$ equals

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$\sqrt{x + 2} \sqrt{5 - x} + 5 \sin^{- 1} \sqrt{\frac{x + 2}{5}} + C$

$\sqrt{x + 2} \sqrt{5 - x} + 5 \sin^{- 1} \sqrt{x + 2} + C$

$\sqrt{x + 2} \sqrt{5 - x} + 7 \sin^{- 1} \sqrt{\frac{x + 2}{7}} + C$

$\sqrt{x + 2} \sqrt{5 - x} + 3 \sin^{- 1} \sqrt{\frac{x + 2}{3}} + C$

answer is B.

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Detailed Solution

Put $2 + x = t^{2},$ so that $d x = 2 t d t,$ then

$I = \int \frac{\sqrt{7 - t^{2}}}{t} (2 t) d t = 2 \int \sqrt{7 - t^{2}} d t$

$= t \sqrt{7 - t^{2}} + 7 \sin^{- 1} (\frac{t}{\sqrt{7}}) + C$

$= \sqrt{x + 2} \sqrt{5 - x} + 7 \sin^{- 1} (\frac{\sqrt{x + 2}}{\sqrt{7}}) + C$

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If $I = \int \sqrt{\frac{5 - x}{2 + x}} d x,$ then $I$ equals