Question

\[\int x^2 \sqrt{x + 2} \text{  dx  }\]

Answer 1

\[\int x^2 \sqrt{x + 2} \text{ dx  }\]
\[\text{Let x + 2 }= t\]
\[ \Rightarrow x = t - 2\]
\[ \Rightarrow dx = dt\]
\[\text{Now,} \int x^2 \sqrt{x + 2} \text{ dx }\]
\[ = \int \left( t - 2 \right)^2 \sqrt{t} \text{ dt }\]
\[ = \int\left( 4^2 - 4t + 4 \right) t^\frac{1}{2} \text{ dt }\]
\[ = \int\left( t^{2 + \frac{1}{2}} - 4 t^{1 + \frac{1}{2}} + 4 t^\frac{1}{2} \right)\text{ dt }\]
\[ = \int\left( t^\frac{5}{2} - 4 t^\frac{3}{2} + 4 t^\frac{1}{2} \right)\text{ dt }\]
\[ = \left[ \frac{t^\frac{5}{2} + 1}{\frac{5}{2} + 1} \right] - 4\left[ \frac{t^\frac{3}{2} + 1}{\frac{3}{2} + 1} \right] + 4\left[ \frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] + C\]
\[ = \frac{2}{7} t^\frac{7}{2} - \frac{8}{5} t^\frac{5}{2} + \frac{8}{3} t^\frac{3}{2} + C\]
\[ = \frac{2}{7} \left( x + 2 \right)^\frac{7}{2} - \frac{8}{5} \left( x + 2 \right)^\frac{5}{2} + \frac{8}{3} \left( x + 2 \right)^\frac{3}{2} + C\]