Question

\[\int x\sqrt{2x + 3} \text{ dx }\]

Answer 1

\[ \text{  Let I }= \int \text{ x}\sqrt{2x + 3} \text{ dx }\]
\[ \text{  Putting 2x + 3 = t }\]
\[ \Rightarrow x = \frac{t - 3}{2}\]
\[ \Rightarrow 2dx = dt\]
\[ \Rightarrow dx = \frac{dt}{2}\]
\[ \therefore I = \frac{1}{2}\int\left( \frac{t - 3}{2} \right) \sqrt{t} \text{ dt }\]
\[ = \frac{1}{4}\int\left( t - 3 \right) \sqrt{t} \text{ dt}\]
\[ = \frac{1}{4}\int\left( t^\frac{3}{2} - 3 t^\frac{1}{2} \right) \text{ dt }\]
\[ = \frac{1}{4}\left[ \frac{t^\frac{3}{2} + 1}{\frac{3}{2} + 1} - 3 \frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] + C\]
\[ = \frac{1}{4} \times \frac{2}{5} t^\frac{5}{2} - \frac{3}{4} \times \frac{2}{3}\text t^\frac{3}{2} + C\]
\[ = \frac{1}{10} \text{ t}^\frac{5}{2} - 2 t^\frac{3}{2} + C\]
\[ = \frac{1}{10} \left( 2x + 3 \right)^\frac{5}{2} - \frac{1}{2} \left( 2x + 3 \right)^\frac{3}{2} + C .........\left[ \because t = 2x + 3 \right]\]
\[ = \frac{1}{10} \left( 2x + 3 \right)^\frac{5}{2} - \frac{1}{2} \left( 2x + 3 \right)^\frac{3}{2} + C\]