Question

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

Answer 1

\[\int\left( 2 x^2 + 3 \right) \sqrt{x + 2} \text{ dx }\]
\[\text{ Let x  }+ 2 = t\]
\[ \Rightarrow x = t - 2\]
\[ \Rightarrow dx = dt\]
\[\int\left[ 2 \left( t - 2 \right)^2 + 3 \right]\sqrt{t}\text{   dt }\]
\[ = \int\left( 2\sqrt{t} \left( t^2 - 4t + 4 \right) + 3\sqrt{t} \right)\text{ dt }\]
\[ = 2\int\left( t^\frac{5}{2} - 4 t^\frac{3}{2} + 4 t^\frac{1}{2} \right) dt + 3\int t^\frac{1}{2} \text{ dt  }\]
\[ = 2\left[ \frac{t^\frac{5}{2} + 1}{\frac{5}{2} + 1} - \frac{4 t^\frac{3}{2} + 1}{\frac{3}{2} + 1} + \frac{4 t^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] + 3\left[ \frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] + C\]
\[ = 2\left[ \frac{2}{7} t^\frac{7}{2} - \frac{8}{5} t^\frac{5}{2} + \frac{8}{3} t^\frac{3}{2} \right] + 2 t^\frac{3}{2} + C\]
\[ = \frac{4}{7} t^\frac{7}{2} - \frac{16}{5} t^\frac{5}{2} + \frac{16}{3} t^\frac{3}{2} + 2 t^\frac{3}{2} + C\]
\[ = \frac{4}{7} t^\frac{7}{2} - \frac{16}{5} t^\frac{5}{2} + \frac{22}{3} t^\frac{3}{2} + C\]
\[ = \frac{4}{7} \left( x + 2 \right)^\frac{7}{2} - \frac{16}{5} \left( x + 2 \right)^\frac{5}{2} + \frac{22}{3} \left( x + 2 \right)^\frac{3}{2} + C\]