Question

Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .

Answer 1

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

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

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

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

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

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

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