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