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$If \int \frac{2 x + 5}{\sqrt{7 - 6 x - x^{2}}} d x = A \sqrt{7 - 6 x - x^{2}} + B \sin^{- 1} (\frac{x + 3}{4}) + C$ $(where C is a constant of integration), then the ordered pair (A, B) is equal to:$

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$(- 2, - 1)$

$(2, - 1)$

$(- 2, 1)$

$(2, 1)$

answer is A.

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Detailed Solution

$Let I = \int \frac{2 x + 5}{\sqrt{7 - 6 x - x^{2}}} d x$

$= \int \frac{2 x + 6 - 1}{\sqrt{7 - 6 x - x^{2}}} d x$

$= - \int \frac{- 2 x - 6}{\sqrt{7 - 6 x - x^{2}}} d x - \int \frac{1}{\sqrt{16 - {(x + 3)}^{2}}} d x$

$∴ I = - 2 \sqrt{7 - 6 x - x^{2}} - \sin^{- 1} \frac{x + 3}{4}$

$= A \sqrt{7 - 6 x - x^{2}} + B \sin^{- 1} \frac{x + 3}{4}$

$\Rightarrow A = - 2, B = - 1$

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