Question

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

Answer 1

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

\[\text{ Also, }4x + 1 = \lambda\frac{d}{dx}\left( x^2 - x - 2 \right) + \mu\]

\[ \Rightarrow 4x + 1 = \lambda\left( 2x - 1 \right) + \mu\]

\[ \Rightarrow 4x + 1 = \left( 2\lambda \right)x + \mu - \lambda\]

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

\[2\lambda = 4\]

\[ \Rightarrow \lambda = 2\]

\[\text{ And }\]

\[\mu - \lambda = 1\]

\[ \Rightarrow \mu - 2 = 1\]

\[ \Rightarrow \mu = 3\]

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

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

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

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

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

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

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

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

\[ = 2 \left[ \frac{t^\frac{3}{2}}{\frac{3}{2}} \right] + \frac{3}{4} \left( 2x - 1 \right) \sqrt{x^2 - x - 2} - \frac{27}{8}\text{ log } \left| \left( x - \frac{1}{2} \right) + \sqrt{x^2 - x - 2} \right| + C\]

\[ = \frac{4}{3} \left( x^2 - x - 2 \right)^\frac{3}{2} + \frac{3}{4} \left( 2x - 1 \right) \sqrt{x^2 - x - 2} - \frac{27}{8}\text{ log }\left| \left( x - \frac{1}{2} \right) + \sqrt{x^2 - x - 2} \right| + C\]