Question

Integrate the following functions.
$\int \frac{1}{\sqrt{7-6x-x^2}}dx.$

Answer 1

Let $I=\int \frac{1}{\sqrt{7-6x-x^2}}dx=\int \frac{1}{\sqrt{7-(x^2+6x)}}dx$
$=\int \frac{1}{\sqrt{7-[x^2+2\times 3x+(3{)}^2-\left(3{)}^2\right]}}dx=\int \frac{1}{\sqrt{7-[x^2+6x+3^2-9]}}dx=\int \frac{1}{\sqrt{7-\left[\right(x+3{)}^2-9]}}dx=\int \frac{1}{\sqrt{4^2-(x+3{)}^2}}dx$
Let $x+3=t\Rightarrow dx=dt$
$\therefore I=\int \frac{1}{\sqrt{4^2-t^2}}dt=si{n}^{-1}\left(\frac{t}{4}\right)+C[\because \int \frac{dx}{\sqrt{a^2-x^2}}=\frac{1}{a}si{n}^{-1}(\frac{x}{a}\left)\right]=si{n}^{-1}\left(\frac{x+3}{4}\right)+C(\because t=x+3)$