Question

$\int x\sqrt{2x+3}dx$

Answer 1

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