Question

\[\int\frac{x}{\sqrt{8 + x - x^2}} dx\]

Answer 1

 

  ` \text{ Let I }=∫  {x   dx}/{\sqrt{8 + x - x^2}} `

\[\text{ Consider }, x = A\frac{d}{dx} \left( 8 + x - x^2 \right) + B\]

\[x = A \left( 1 - 2x \right) + B\]

\[x = \left( - 2A \right) x + A + B\]

\[\text{ Equating Coefficients of like terms }\]

\[ - 2A = 1\]

\[ \Rightarrow A = - \frac{1}{2}\]

\[\text{ And }\]

\[A + B = 0\]

\[ \Rightarrow - \frac{1}{2} + B = 0\]

\[ \Rightarrow B = \frac{1}{2}\]

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

\[\text{ Then }, \]

\[I = - \frac{1}{2}\int\frac{\left( 1 - 2x \right) dx}{\sqrt{8 + x - x^2}} + \frac{1}{2}\int\frac{dx}{\sqrt{8 + x - x^2}}\]

\[ = - \frac{1}{2}\int\frac{\left( 1 - 2x \right) dx}{\sqrt{8 + x - x^2}} + \frac{1}{2}\int\frac{dx}{\sqrt{8 - \left( x^2 - x \right)}}\]

\[ = - \frac{1}{2}\int\frac{\left( 1 - 2x \right) dx}{\sqrt{8 + x - x^2}} + \frac{1}{2}\int\frac{dx}{\sqrt{8 - \left( x^2 - x + \frac{1}{4} - \frac{1}{4} \right)}}\]

\[ = - \frac{1}{2}\int\frac{\left( 1 - 2x \right) dx}{\sqrt{8 + x - x^2}} + \frac{1}{2}\int\frac{dx}{\sqrt{8 + \frac{1}{4} - \left( x - \frac{1}{2} \right)^2}}\]

\[ = - \frac{1}{2}\int\frac{\left( 1 - 2x \right) dx}{\sqrt{8 + x - x^2}} + \frac{1}{2}\int\frac{dx}{\sqrt{\left( \frac{\sqrt{33}}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2}}\]

\[\text{ let 8 + x - x^2 = t }\]

\[ \Rightarrow \left( 1 - 2x \right) dx = dt\]

\[ \therefore I = - \frac{1}{2}\int\frac{dt}{\sqrt{t}} + \frac{1}{2}\int\frac{dx}{\sqrt{\left( \frac{\sqrt{33}}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2}}\]

\[ = - \frac{1}{2} \times 2\sqrt{t} + \frac{1}{2} \sin^{- 1} \left( \frac{x - \frac{1}{2}}{\frac{\sqrt{33}}{2}} \right) + C\]

\[ = - \sqrt{t} + \frac{1}{2} \sin^{- 1} \left( \frac{2x - 1}{\sqrt{33}} \right) + C\]

\[ = - \sqrt{8 + x - x^2} + \frac{1}{2} \sin^{- 1} \left( \frac{2x - 1}{\sqrt{33}} \right) + C\]