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Question:
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$$
\int \frac{2 x+5}{\sqrt{7-6 x-x^2}} d x=A \sqrt{7-6 x-x^2}+B \sin ^{-1}\left(\frac{x+3}{4}\right)+C
$$
(where $C$ is a constant of integration), then the ordered pair $(A, B)$ is equal to
Options:
\int \frac{2 x+5}{\sqrt{7-6 x-x^2}} d x=A \sqrt{7-6 x-x^2}+B \sin ^{-1}\left(\frac{x+3}{4}\right)+C
$$
(where $C$ is a constant of integration), then the ordered pair $(A, B)$ is equal to
Solution:
2601 Upvotes
Verified Answer
The correct answer is:
$(-2,-1)$
$(-2,-1)$
$\because 7-6 x-x^2=16-(x+3)^2$
and $\frac{d}{d x}\left(7-6 x-x^2\right)=-2 x-6$
So, $\int \frac{2 x+5}{\sqrt{7-6 x-x^2}} d x=\int \frac{2 x+6}{\sqrt{7-6 x-x^2}} d x$
$$
-\int \frac{1}{\sqrt{16-(x+3)^2}} d x
$$
$$
=-2 \sqrt{7-6 x-x^2}-\sin ^{-1}\left(\frac{x+3}{4}\right)+C
$$
Therefore, $A=-2, \& B=-1$.
and $\frac{d}{d x}\left(7-6 x-x^2\right)=-2 x-6$
So, $\int \frac{2 x+5}{\sqrt{7-6 x-x^2}} d x=\int \frac{2 x+6}{\sqrt{7-6 x-x^2}} d x$
$$
-\int \frac{1}{\sqrt{16-(x+3)^2}} d x
$$
$$
=-2 \sqrt{7-6 x-x^2}-\sin ^{-1}\left(\frac{x+3}{4}\right)+C
$$
Therefore, $A=-2, \& B=-1$.
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