∫dx(1+x)8+7x−x2 Put 1+x=1t dx=-1t2dt I=∫ -dt8t2+7(t-t2)-(1-t)2 =∫ -dt9t-1 =19∫-9dt9t-1+c =19(2)9t-1+c =2991+x-1+c =298-x1+x+c

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$\int \frac{d x}{(1 + x) \sqrt{8 + 7 x - x^{2}}} =$

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$- \frac{1}{9} \sqrt{\frac{1 + x}{8 - x}} + c$

$\frac{2}{9} \sqrt{\frac{8 - x}{1 + x}} + c$

$- \frac{2}{9} \sqrt{\frac{8 - x}{1 + x}} + c$

$- \frac{2}{9} \sqrt{\frac{1 + x}{8 - x}} + c$

answer is D.

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Detailed Solution

$\int \frac{dx}{(1 + x) \sqrt{8 + 7 x - x^{2}}} Put 1 + x = \frac{1}{t} dx = \frac{- 1}{t^{2}} dt I = \int \frac{- dt}{\sqrt{8 t^{2} + 7 (t - t^{2}) - {(1 - t)}^{2}}}$

$= \int \frac{- dt}{\sqrt{9 t - 1}} = \frac{1}{9} \int \frac{- 9 dt}{\sqrt{9 t - 1}} + c = \frac{1}{9} (2) \sqrt{9 t - 1} + c = \frac{2}{9} \sqrt{\frac{9}{1 + x} - 1} + c = \frac{2}{9} \sqrt{\frac{8 - x}{1 + x}} + c$