Question

If $\int \frac{2x + 5}{\sqrt{7 - 6x - x^2}} dx = A \sqrt{7 - 6x - 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 :

• A. (2, 1)
• B. (-2, -1)
• C. (-2, 1)
• D. (2, -1)

Answer 1

Answer: (B)

Given: $\int \frac{2 x+5}{\sqrt{7-6 x-x^{2}}} d x=\int \frac{a \frac{d}{d x}\left(7-6 x-x^{2}\right)+b}{\sqrt{7-6 x-x^{2}}} d x$
$2 x+5=a(-6-2 x)+b$
Now comparing $x$ coefficient and constant coefficient, we get
$5=-6 a+b $ and $-2 a=2 \Rightarrow a=-1 $
$5=6+b \Rightarrow b=-1 $
$-\int \frac{(-6-2 x)}{\sqrt{7-6 x-x^{2}}} d x-\int \frac{d x}{\sqrt{16-(x+3)^{2}}}=-2 \sqrt{7-6 x-x^{2}}-\sin ^{-1}\left(\frac{x+3}{4}\right)+c$
$A=-2, B=-1$