Question

\[\int\frac{1}{\sqrt{3 - 2x - x^2}} \text{ dx}\]

Answer 1

\[\text{ Let I} = \int\frac{1}{\sqrt{3 - 2x - x^2}}dx\]

\[ = \int\frac{1}{\sqrt{3 - \left( x^2 + 2x + 1 - 1 \right)}}dx\]

\[ = \int\frac{1}{\sqrt{4 - \left( x + 1 \right)^2}}dx\]

\[\text{ Putting} \left( x + 1 \right) = t\]

\[ \Rightarrow dx = dt\]

\[ \therefore I = \int\frac{dt}{\sqrt{2^2 - t^2}}\]

\[ = \sin^{- 1} \left( \frac{t}{2} \right) + C .................\left[ \because \int \frac{1}{\sqrt{a^2 - x^2}}dx = \sin^{- 1} \frac{x}{a} + C \right]\]

\[ = \sin^{- 1} \left( \frac{x + 1}{2} \right) + C .....................\left[ \because t = \left( x + 1 \right) \right]\]