Question

\[\int\left( 2x - 5 \right) \sqrt{2 + 3x - x^2} \text{  dx }\]

Answer 1

\[\text{ Let I }= \int \left( 2x - 5 \right) \sqrt{2 + 3x - x^2} \text{  dx }\]

\[\text{ Also,} 2x - 5 = \lambda\frac{d}{dx}\left( 2 + 3x - x^2 \right) + \mu\]

\[ \Rightarrow 2x - 5 = \lambda\left( - 2x + 3 \right) + \mu\]

\[ \Rightarrow 2x - 5 = \left( - 2\lambda \right)x + 3\lambda + \mu\]

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

\[ - 2\lambda = 2\]

\[ \Rightarrow \lambda = - 1\]

\[\text{ And }\]

\[3\lambda + \mu = - 5\]

\[ \Rightarrow 3\left( - 1 \right) + \mu = - 5\]

\[ \Rightarrow \mu = - 5 + 3\]

\[ \Rightarrow \mu = - 2\]

\[ \therefore 2x - 5 = - 1\left( - 2x + 3 \right) - 2\]

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

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

\[ = - I_1 - 2 I_2 . . . . . \left( 1 \right)\]

\[ I_1 = \int\left( - 2x + 3 \right) \sqrt{2 + 3x - x^2} \text{  dx }\]

\[\text{ Let } 2 + 3x - x^2 = t\]

\[ \Rightarrow \left( - 2x + 3 \right)dx = dt\]

\[ \therefore I_1 = \int t^\frac{1}{2} dt\]

\[ = \frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1}\]

\[ = \frac{2}{3} t^\frac{3}{2} \]

\[ = \frac{2}{3} \left( 2 + 3x - x^2 \right)^\frac{3}{2} . . . . . \left( 2 \right)\]

\[\text{ And I}_2 = \int \sqrt{2 + 3x - x^2} \text{  dx }\]

\[ I_2 = \int \sqrt{2 - \left( x^2 - 3x \right)} \text{  dx }\]

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

\[ = \int\sqrt{2 + \frac{9}{4} - \left( x - \frac{3}{2} \right)^2} dx\]

\[ = \int \sqrt{\left( \frac{\sqrt{17}}{2} \right)^2 - \left( x - \frac{3}{2} \right)^2} \text{  dx }\]

\[ = \frac{x - \frac{3}{2}}{2} \sqrt{\left( \frac{\sqrt{17}}{2} \right)^2 - \left( x - \frac{3}{2} \right)^2} + \frac{\left( \frac{\sqrt{17}}{2} \right)^2}{2} \sin^{- 1} \left( \frac{x - \frac{3}{2}}{\frac{\sqrt{17}}{2}} \right)\]

\[ = \frac{2x - 3}{4}\sqrt{2 + 3x - x^2} + \frac{17}{8} \sin^{- 1} \left( \frac{2x - 3}{\sqrt{17}} \right) . . . . . \left( 3 \right)\]

\[\text{ From eq }\left( 1 \right), \left( 2 \right) \text{ and } \left( 3 \right) \text{ we have}\]

\[I = - \frac{2}{3} \left( 2 + 3x - x^2 \right)^\frac{3}{2} - \frac{\left( 2x - 3 \right)}{2}\sqrt{2 + 3x - x^2} - \frac{17}{4} \sin^{- 1} \left( \frac{2x - 3}{\sqrt{17}} \right) + C\]