Mathematics - Indefinite Integration Question with Solution | TestHub

$\because 7-6x-x^2=16-(x+3{)}^2$ and $\frac{d}{dx}\left(7-6x-x^2\right)=-2x-6$ So, $\int \frac{2x+5}{\sqrt{7-6x-x^2}}dx=\int \frac{2x+6}{\sqrt{7-6x-x^2}}dx$ $$-\int \frac{1}{\sqrt{16-(x+3{)}^2}}dx$$ $$=-2\sqrt{7-6x-x^2}-{\sin }^{-1}\left(\frac{x+3}{4}\right)+C$$ Therefore, $A=-2,\&B=-1$.