Question

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

Answer 1

\[\text{ Let I }= \int x\sqrt{1 + x - x^2}\text{ dx}\]
\[\text{ and  let x }= A\frac{d}{dx}\left( 1 + x - x^2 \right) + B\]
\[ \Rightarrow x = A \left( - 2x + 1 \right) + B\]
\[\text{By equating the coefficients of like terms we get}, \]
\[x = \left( - 2A \right) x\]
\[ \Rightarrow A = - \frac{1}{2}\]
\[\text{  and   A + B = 0 }\]
\[ \Rightarrow B = \frac{1}{2}\]
\[\text{By substituting the values of A and B in eq (1) we get}, \]
\[I = \int\left[ - \frac{1}{2} \left( - 2x + 1 \right) + \frac{1}{2} \right] \sqrt{1 + x - x^2} \text{ dx }\]
\[ = - \frac{1}{2}\int\left( - 2x + 1 \right) \sqrt{1 + x - x^2}dx + \frac{1}{2} \sqrt{1 + x - x^2}\text{ dx }\]
\[\text{ Putting  1 + x - x^2 = t}\]
\[ \Rightarrow \left( - 2x + 1 \right) \text{ dx }= dt\]
\[ \therefore I = - \frac{1}{2}\int\sqrt{t} \cdot dt + \frac{1}{2}\int\sqrt{1 + x - x^2} \text{ dx }\]
\[ = - \frac{1}{2}\int\sqrt{t} \text{ dt} + \frac{1}{2}\int\sqrt{1 - \left( x^2 - x \right)} \text{ dx }\]
\[ = - \frac{1}{2}\int t^\frac{1}{2} \cdot dt + \frac{1}{2}\int\sqrt{1 - \left( x^2 - x + \frac{1}{4} - \frac{1}{4} \right)}\text{ dx }\]
\[ = - \frac{1}{2}\left[ \frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] + \frac{1}{2}\int\sqrt{1 - \left( x - \frac{1}{2} \right)^2 + \frac{1}{4}}\text{ dx }\]
\[ = - \frac{1}{2} \times \frac{2}{3} t^\frac{3}{2} + \frac{1}{2}\int\sqrt{\left( \frac{\sqrt{5}}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2}dx\]
\[ = - \frac{1}{3} t^\frac{3}{2} + \frac{1}{2}\left[ \left( \frac{x - \frac{1}{2}}{2} \right) \sqrt{\left( \frac{\sqrt{5}}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2} + \frac{\left( \frac{\sqrt{5}}{2} \right)^2}{2} \text{ sin}^{- 1} \left( \frac{x - \frac{1}{2}}{\frac{\sqrt{5}}{2}} \right) \right] + C ......................\left[ \because \int\sqrt{a^2 - x^2}dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{1}{2} a^2 \sin^{- 1} \frac{x}{a} + C \right]\]
\[ = - \frac{1}{3} \left( 1 + x - x^2 \right)^\frac{3}{2} + \frac{1}{2}\left[ \left( \frac{2x - 1}{4} \right) \sqrt{1 + x - x^2} + \frac{5}{8} \text{ sin}^{- 1} \left( \frac{2x - 1}{\sqrt{5}} \right) \right] + C\]
\[ = \frac{- \left( 1 + x - x^2 \right)\sqrt{1 + x - x^2}}{3} + \frac{\left( 2x - 1 \right)}{8} \sqrt{1 + x - x^2} + \frac{5}{16} \text{ sin}^{- 1} \left( \frac{2x - 1}{\sqrt{5}} \right) + C\]
\[ = \sqrt{1 + x - x^2} \left[ \frac{- \left( 1 + x - x^2 \right)}{3} + \frac{2x - 1}{8} \right] + \frac{5}{16} \text{ sin}^{- 1} \left( \frac{2x - 1}{\sqrt{5}} \right) + C\]
\[ = \sqrt{1 + x - x^2} \left[ \frac{- 8 - 8x + 8 x^2 + 6x - 3}{24} \right] + \frac{5}{16}\text{ sin}^{- 1} \left( \frac{2x - 1}{\sqrt{5}} \right) + C\]
\[ = \sqrt{1 + x - x^2} \left[ \frac{8 x^2 - 2x - 11}{24} \right] + \frac{5}{16} \text{ sin}^{- 1} \left( \frac{2x - 1}{\sqrt{5}} \right) + C\]