Question

$\int \frac{1}{\sqrt{3-2x-x^2}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{\sqrt{3-2x-x^2}}dx \\ =\int \frac{1}{\sqrt{3-\left(x^2+2x+1-1\right)}}dx \\ =\int \frac{1}{\sqrt{4-{\left(x+1\right)}^2}}dx \\ \mathrm{Putting}\left(x+1\right)=t \\ \Rightarrow dx=dt \\ \therefore I=\int \frac{dt}{\sqrt{2^2-t^2}} \\ ={\sin }^{-1}\left(\frac{t}{2}\right)+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because \int \frac{1}{\sqrt{a^{\mathit{2}}\mathit{-}{x}^{\mathit{2}}}}dx\mathit{=}{\sin }^{-1}\frac{x}{a}+C\right] \\ ={\sin }^{-1}\left(\frac{x+1}{2}\right)+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because \mathit{}t\mathit{}=\left(\mathit{}x+1\right)\right] \end{gathered}$$