Question

$\int \frac{2-3x}{\sqrt{1+3x}}dx$

Answer 1

$\mathrm{Let}I=\int \left(\frac{2-3x}{\sqrt{1+3x}}\right)dx$
Putting 1 + 3x = t
⇒ 3x = t – 1

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

$$\begin{gathered} \therefore I=\int \left(\frac{2-\left(t-1\right)}{\sqrt{t}}\right)dt \\ =\int \left(\frac{3-t}{\sqrt{t}}\right)dt \\ =\int \left(3t^{-\frac{1}{2}}-t^{\frac{1}{2}}\right)dt \\ =3\int t^{-\frac{1}{2}}dt-\int t^{\frac{1}{2}}dt \\ =3\left[\frac{t^{{\displaystyle \frac{-1}{2}}+1}}{-{\displaystyle \frac{1}{2}}+1}\right]-\left[\frac{t^{{\displaystyle \frac{1}{2}}+1}}{{\displaystyle \frac{1}{2}}+1}\right]+\mathrm{C} \\ =6\sqrt{t}-\frac{2}{3}{t}^{\frac{3}{2}}+\mathrm{C} \\ =2\sqrt{t}\left(3-\frac{t}{3}\right)+\mathrm{C} \\ =2\sqrt{t}\left(\frac{9-t}{3}\right)+\mathrm{C}\left[\because t=1+3x\right] \\ =\frac{2}{3}\sqrt{1+3x}\left\{\frac{9-\left(1+3x\right)}{3}\right\}+\mathrm{C} \\ =\frac{2}{3\times 3}\sqrt{1+3x}\left(8-3x\right)+\mathrm{C} \\ =\frac{2}{9}\left(8-3x\right)\sqrt{1+3x}+\mathrm{C} \end{gathered}$$