Question

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

Answer 1

$$\begin{gathered} \int \left(\frac{1}{2-3x}+\frac{1}{\sqrt{3x-2}}\right)dx \\ =\int \frac{dx}{2-3x}+\int {\left(3x-2\right)}^{-\frac{1}{2}}dx \\ =\frac{\ln \left(2-3x\right)}{-3}+\left[\frac{{\left(3x-2\right)}^{-{\displaystyle \frac{1}{2}}+1}}{3\left(-{\displaystyle \frac{1}{2}}+1\right)}\right]+\mathrm{C} \\ =\frac{\ln \left(2-3x\right)}{-3}+\frac{2}{3}{\left(3x-2\right)}^{\frac{1}{2}}+\mathrm{C} \\ =-\frac{1}{3}\ln \left(2-3\mathrm{x}\right)+\frac{2}{3}\sqrt{3\mathrm{x}-2}+\mathrm{C} \end{gathered}$$