Question

$\int \left(x+2\right)\sqrt{3x+5}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(x+2\right)\sqrt{3x+5}dx \\ \mathrm{Putting}3x+5=t \\ \Rightarrow x=\frac{t-5}{3} \end{gathered}$$

$$\begin{gathered} \Rightarrow 3dx=dt \\ \Rightarrow dx=\frac{dt}{3} \end{gathered}$$

$$\begin{gathered} \therefore I=\int \left(\frac{t-5}{3}+2\right)\sqrt{t}\frac{dt}{3} \\ =\frac{1}{3}\int \left(\frac{t-5+6}{3}\right)\sqrt{t}dt \\ =\frac{1}{9}\int \left(t+1\right)\sqrt{t}dt \\ =\frac{1}{9}\int \left(t^{\frac{3}{2}}+t^{\frac{1}{2}}\right)dt \\ =\frac{1}{9}\left[\frac{t^{{\displaystyle \frac{3}{2}}+1}}{{\displaystyle \frac{3}{2}}+1}+\frac{t^{{\displaystyle \frac{1}{2}}+1}}{{\displaystyle \frac{1}{2}}+1}\right]+\mathrm{C} \\ =\frac{1}{9}\left[\frac{2}{5}{t}^{\frac{5}{2}}+\frac{2}{3}{t}^{\frac{3}{2}}\right]+\mathrm{C} \\ =\frac{1}{9}\left[\frac{2}{5}{\left(3x+5\right)}^{\frac{5}{2}}+\frac{2}{3}{\left(3x+5\right)}^{\frac{3}{2}}\right]+\mathrm{C}\left[\because t=3x+5\right] \\ =\frac{2}{9}\left[{\left(3x+5\right)}^{\frac{3}{2}}\left\{\frac{3x+5}{5}+\frac{1}{3}\right\}\right]+\mathrm{C} \\ =\frac{2}{9}\left[{\left(3x+5\right)}^{\frac{3}{2}}\left\{\frac{9x+15+5}{15}\right\}\right]+\mathrm{C} \\ =\frac{2}{9}\left[{\left(3x+5\right)}^{\frac{3}{2}}\left\{\frac{9x+20}{15}\right\}\right]+\mathrm{C} \\ =\frac{2}{135}{\left(3x+5\right)}^{\frac{3}{2}}\left(9x+20\right)+\mathrm{C} \end{gathered}$$