Question

\[\int x\sqrt{x^2 + x} \text{  dx }\]

Answer 1

\[\text{ Let I } = \int x\sqrt{x^2 + x}dx\]
\[\text{ Also, }x = \lambda\frac{d}{dx}\left( x^2 + x \right) + \mu\]
\[ \Rightarrow x = \lambda\left( 2x + 1 \right) + \mu\]
\[ \Rightarrow x = \left( 2\lambda \right)x + \lambda + \mu\]
\[\text{Equating coefficient of like terms}\]
\[2\lambda = 1\]
\[ \Rightarrow \lambda = \frac{1}{2}\]
\[\text{ And }\]
\[\lambda + \mu = 0\]
\[ \Rightarrow \mu = - \frac{1}{2}\]
\[ \therefore I = \int \left[ \frac{1}{2}\left( 2x + 1 \right) - \frac{1}{2} \right] \sqrt{x^2 + x}dx\]
\[ = \frac{1}{2}\int\left( 2x + 1 \right) \sqrt{x^2 + x}dx - \frac{1}{2}\int\sqrt{x^2 + x}dx\]
\[ = \frac{1}{2}\int \left( 2x + 1 \right) \sqrt{x^2 + x}dx - \frac{1}{2}\int\sqrt{x^2 + x + \frac{1}{4} - \frac{1}{4}}dx\]
\[ = \frac{1}{2}\int\left( 2x + 1 \right) \sqrt{x^2 + x} \text{  dx }- \frac{1}{2}\int\sqrt{\left( x + \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}\text{  dx }\]
\[\text{ Let x}^2 + x = t\]
\[ \Rightarrow \left( 2x + 1 \right)dx = dt\]
\[\text{ Then,} \]
\[I = \frac{1}{2}\int\sqrt{t} \text{ dt }- \frac{1}{2}\left[ \frac{x + \frac{1}{2}}{2} \sqrt{x^2 + x} - \frac{1}{8}\text{ log }\left| \left( x + \frac{1}{2} \right) + \sqrt{x^2 + x} \right| \right] + C\]
\[ = \frac{1}{2} \times \frac{2}{3} t^\frac{3}{2} - \left( \frac{2x + 1}{8} \right) \sqrt{x^2 + x} + \frac{1}{16}\text{ log } \left| \left( x + \frac{1}{2} \right) + \sqrt{x^2} + x \right| + C\]
\[ = \frac{1}{3} \left( x^2 + x \right)^\frac{3}{2} - \left( \frac{2x + 1}{8} \right) \sqrt{x^2 + x} + \frac{1}{16}\text{ log } \left| \left( x + \frac{1}{2} \right) + \sqrt{x^2} + x \right| + C\]