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 orderd 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 C $(- 2, 1)$
Let $I = \int \frac{2 x + 5}{\sqrt{7 - 6 x - x^{2}}} d x$
$I = \int \frac{2 x + 6}{\sqrt{7 - 6 x - x^{2}}} d x - \int \frac{1}{\sqrt{7 - 6 x - x^{2}}} d x$
Let $I = I_{1} + I_{2}$
So, $I_{1} = \int \frac{2 x + 6}{\sqrt{7 - 6 x - x^{2}}} d x$
Put $7 - 6 x - x^{2} = t^{2} \Rightarrow - (2 x + 6) d x = 2 t d t$
$\Rightarrow I_{1} = \int - 2 d t = - 2 t = - 2 \sqrt{7 - 6 x - x^{2}} + c_{1}$
Now, $I_{2} = \int \frac{1}{\sqrt{7 - 6 x - x^{2}}} d x$
$= \int \frac{1}{\sqrt{4^{2} - (x + 3)^{2}}} d x$
$\Rightarrow I_{2} = {sin}^{- 1} (\frac{x + 3}{4}) + c_{2}$
therefore, $I = - 2 \sqrt{7 - 6 x - x^{2}} + {sin}^{- 1} (\frac{x + 3}{4}) + C$
Hence, $A = - 2$ and $B = 1$

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