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प्रश्न
\[\int\frac{x}{\sqrt{8 + x - x^2}} dx\]
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उत्तर
` \text{ Let I }=∫ {x dx}/{\sqrt{8 + x - x^2}} `
\[\text{ Consider }, x = A\frac{d}{dx} \left( 8 + x - x^2 \right) + B\]
\[x = A \left( 1 - 2x \right) + B\]
\[x = \left( - 2A \right) x + A + B\]
\[\text{ Equating Coefficients of like terms }\]
\[ - 2A = 1\]
\[ \Rightarrow A = - \frac{1}{2}\]
\[\text{ And }\]
\[A + B = 0\]
\[ \Rightarrow - \frac{1}{2} + B = 0\]
\[ \Rightarrow B = \frac{1}{2}\]
\[ \therefore x = - \frac{1}{2} \left( 1 - 2x \right) + \frac{1}{2}\]
\[\text{ Then }, \]
\[I = - \frac{1}{2}\int\frac{\left( 1 - 2x \right) dx}{\sqrt{8 + x - x^2}} + \frac{1}{2}\int\frac{dx}{\sqrt{8 + x - x^2}}\]
\[ = - \frac{1}{2}\int\frac{\left( 1 - 2x \right) dx}{\sqrt{8 + x - x^2}} + \frac{1}{2}\int\frac{dx}{\sqrt{8 - \left( x^2 - x \right)}}\]
\[ = - \frac{1}{2}\int\frac{\left( 1 - 2x \right) dx}{\sqrt{8 + x - x^2}} + \frac{1}{2}\int\frac{dx}{\sqrt{8 - \left( x^2 - x + \frac{1}{4} - \frac{1}{4} \right)}}\]
\[ = - \frac{1}{2}\int\frac{\left( 1 - 2x \right) dx}{\sqrt{8 + x - x^2}} + \frac{1}{2}\int\frac{dx}{\sqrt{8 + \frac{1}{4} - \left( x - \frac{1}{2} \right)^2}}\]
\[ = - \frac{1}{2}\int\frac{\left( 1 - 2x \right) dx}{\sqrt{8 + x - x^2}} + \frac{1}{2}\int\frac{dx}{\sqrt{\left( \frac{\sqrt{33}}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2}}\]
\[\text{ let 8 + x - x^2 = t }\]
\[ \Rightarrow \left( 1 - 2x \right) dx = dt\]
\[ \therefore I = - \frac{1}{2}\int\frac{dt}{\sqrt{t}} + \frac{1}{2}\int\frac{dx}{\sqrt{\left( \frac{\sqrt{33}}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2}}\]
\[ = - \frac{1}{2} \times 2\sqrt{t} + \frac{1}{2} \sin^{- 1} \left( \frac{x - \frac{1}{2}}{\frac{\sqrt{33}}{2}} \right) + C\]
\[ = - \sqrt{t} + \frac{1}{2} \sin^{- 1} \left( \frac{2x - 1}{\sqrt{33}} \right) + C\]
\[ = - \sqrt{8 + x - x^2} + \frac{1}{2} \sin^{- 1} \left( \frac{2x - 1}{\sqrt{33}} \right) + C\]
