Question

$\int \frac{\sqrt{a}-\sqrt{x}}{1-\sqrt{ax}}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{\sqrt{a}-\sqrt{x}}{1-\sqrt{ax}}dx \\ I=\frac{1}{\sqrt{a}}\int \frac{1+a-1-\sqrt{ax}}{1-\sqrt{ax}}dx \\ I=\frac{1}{\sqrt{a}}\int \frac{1-\sqrt{ax}}{1-\sqrt{ax}}dx+\frac{1}{\sqrt{a}}\int \frac{a-1}{1-\sqrt{ax}}dx \\ I=\frac{1}{\sqrt{a}}\int dx+\frac{a-1}{\sqrt{a}}\int \frac{1}{1-\sqrt{ax}}dx \\ I=\frac{1}{\sqrt{a}}x+\frac{a-1}{\sqrt{a}}\int \frac{1}{1-\sqrt{ax}}dx \\ \mathrm{Let}, \\ {I}_1=\int \frac{1}{1-\sqrt{ax}}dx \\ \mathrm{Put}ax=z^2 \\ \Rightarrow adx=2zdz \\ {I}_1=\frac{1}{a}\int \frac{2z}{1-z}dz \\ {I}_1=\frac{1}{a}\int \frac{2z-2+2}{1-z}dz \\ {I}_1=\frac{1}{a}\int \frac{2z-2}{1-z}dz+\frac{1}{a}\int \frac{2}{1-z}dz \\ {I}_1=\frac{-2}{a}\int \frac{1-z}{1-z}dz+\frac{1}{a}\int \frac{2}{1-z}dz \\ {I}_1=\frac{-2}{a}\int dz+\frac{1}{a}\int \frac{2}{1-z}dz \\ {I}_1=\frac{-2}{a}z-\frac{2}{a}\log \left|1-z\right|+C_1 \\ {I}_1=\frac{-2\sqrt{ax}}{a}-\frac{2}{a}\log \left|1-\sqrt{ax}\right|+C_1 \\ I=\frac{1}{\sqrt{a}}x+\frac{a-1}{\sqrt{a}}\left(\frac{-2\sqrt{ax}}{a}-\frac{2}{a}\log \left|1-\sqrt{ax}\right|\right)+C \end{gathered}$$

Note: The answer in indefinite integration may vary depending on the integral constant.