Question

\[\int\left( x - 2 \right) \sqrt{2 x^2 - 6x + 5} \text{  dx }\]

Answer 1

\[\text{ Let I }= \int \left( x - 2 \right) \sqrt{2 x^2 - 6x + 5} \text{  dx }\]

\[\text{ Also, } x - 2 = \lambda\frac{d}{dx}\left( 2 x^2 - 6x + 5 \right) + \mu\]

\[ \Rightarrow x - 2 = \left( 4\lambda \right)x + \mu - 6\lambda\]

\[\text{Equating the coefficient of like terms}\]

\[4\lambda = 1\]

\[ \Rightarrow \lambda = \frac{1}{4}\]

\[\text{ And }\]

\[\mu - 6\lambda = - 2\]

\[ \Rightarrow \mu - 6 \times \frac{1}{4} = - 2\]

\[ \Rightarrow \mu = - 2 + \frac{3}{2} = - \frac{1}{2}\]

\[ \therefore I = \int \left[ \frac{1}{4}\left( 4x - 6 \right) - \frac{1}{2} \right] \sqrt{2 x^2 - 6x + 5} \text{  dx }\]

\[ = \frac{1}{4} \int \left( 4x - 6 \right) \sqrt{2 x^2 - 6x + 5} dx - \frac{1}{2}\int\sqrt{2 x^2 - 6x + 5} \text{  dx }\]

\[\text{ Let 2 x }^2 - 6x + 5 = t\]

\[ \Rightarrow \left( 4x - 6 \right)dx = dt\]

\[ \therefore I = \frac{1}{4}\int t^\frac{1}{2} \text{ dt }- \frac{1}{2}\int\sqrt{2\left( x^2 - 3x + \frac{5}{2} \right)}\text{  dx }\]

\[ = \frac{1}{4}\int t^\frac{1}{2} - \frac{\sqrt{2}}{2}\int\sqrt{x^2 - 3x + \left( \frac{3}{2} \right)^2 - \left( \frac{3}{2} \right)^2 + \frac{5}{2}} \text{  dx }\]

\[ = \frac{1}{4}\left[ \frac{t^\frac{3}{2}}{\frac{3}{2}} \right] - \frac{1}{\sqrt{2}}\int \sqrt{\left( x - \frac{3}{2} \right)^2 - \frac{9}{4} + \frac{5}{2}} \text{  dx }\]

\[ = \frac{1}{6} t^\frac{3}{2} - \frac{1}{\sqrt{2}} \int \sqrt{\left( x - \frac{3}{2} \right)^2 - \frac{9 + 10}{4}} \text{  dx }\]

\[ = \frac{1}{6} t^\frac{3}{2} - \frac{1}{\sqrt{2}}\int\sqrt{\left( x - \frac{3}{2} \right)^2 + \left( \frac{1}{2} \right)^2} \text{  dx }\]

\[ = \frac{1}{6} \left( 2 x^2 - 6x + 5 \right)^\frac{3}{2} - \frac{1}{\sqrt{2}} \left[ \left( \frac{x - \frac{3}{2}}{2} \right) \sqrt{\left( x - \frac{3}{2} \right)^2 + \left( \frac{1}{2} \right)^2} + \frac{1}{8}\text{ log }\left| \left( x - \frac{3}{2} \right) + \sqrt{x^2 - 3x + \frac{5}{2}} \right| \right] + C\]

\[ = \frac{1}{6} \left( 2 x^2 - 6x + 5 \right)^\frac{3}{2} - \frac{1}{\sqrt{2}} \left[ \frac{2x - 3}{4} \sqrt{x^2 - 3x + \frac{5}{2}} + \frac{1}{8}\text{ log} \left| \frac{2x - 3}{2} + \sqrt{x^2 - 3x + \frac{5}{2}} \right| \right] + C\]