Question

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

Answer 1

\[\int\sqrt{1 + 2x - 3 x^2}dx\]
\[ = \sqrt{3}\int\sqrt{\frac{1}{3} + \frac{2}{3}x - x^2}dx\]
\[ = \sqrt{3}\int\sqrt{\frac{1}{3} - \left( x^2 - \frac{2}{3}x \right)}dx\]
\[ = \sqrt{3}\int\sqrt{\frac{1}{3} - \left\{ x^2 - \frac{2}{3}x + \left( \frac{1}{3} \right)^2 - \left( \frac{1}{3} \right)^2 \right\}}dx\]
\[ = \sqrt{3}\int\sqrt{\frac{1}{3} + \frac{1}{9} - \left( x - \frac{1}{3} \right)^2}dx\]
\[ = \sqrt{3}\int\sqrt{\frac{4}{9} - \left( x - \frac{1}{3} \right)^2}dx\]
\[ = \sqrt{3}\int\sqrt{\left( \frac{2}{3} \right)^2 - \left( x - \frac{1}{3} \right)^2}dx\]
\[ = \sqrt{3} \left[ \frac{\left( x - \frac{1}{3} \right)}{2} \sqrt{\left( \frac{2}{3} \right)^2 - \left( x - \frac{1}{3} \right)^2} + \frac{\left( \frac{2}{3} \right)^2}{2} \sin^{- 1} \left( \frac{x - \frac{1}{3}}{\frac{2}{3}} \right) \right] + C ................................\left[ \because \int\sqrt{a^2 - x^2}dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{1}{2} a^2 \sin^{- 1} \frac{x}{a} + C \right]\]
\[ = \left( \frac{3x - 1}{6} \right) \sqrt{1 + 2x - 3 x^2} + \frac{2\sqrt{3}}{9} \sin^{- 1} \left( \frac{3x - 1}{2} \right) + C\]