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

(- 2, - 1)

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(2, - 1)

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(- 2, 1)

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(2, 1)

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Solution

The correct option is D $(- 2, - 1)$
Note that $7 - 6 x - x^{2} = 16 - (x + 3)^{2}$ and $\frac{d}{d x} (7 - 6 x - x^{2}) = - 2 x - 6$

So, we have

$\int \frac{2 x + 5}{\sqrt{7 - 6 x - x^{2}}} d x = \int \frac{2 x + 6}{\sqrt{7 - 6 x - x^{2}}} d x - \int \frac{1}{\sqrt{16 - (x + 3)^{2}}} d x = - 2 \sqrt{7 - 6 x - x^{2}} - {sin}^{- 1} (\frac{x + 3}{4}) + C$

So, we have $A = - 2, B = - 1$ .

Thus option A is the correct answer.

Suggest Corrections

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