Question

\[\int\frac{\sqrt{a} - \sqrt{x}}{1 - \sqrt{ax}}\text{  dx }\]

Answer 1

\[\text{ We  have,} \]

\[I = \int\frac{\sqrt{a} - \sqrt{x}}{1 - \sqrt{ax}} \text{ dx }\]

\[I = \frac{1}{\sqrt{a}}\int\frac{1 + a - 1 - \sqrt{ax}}{1 - \sqrt{ax}} \text{ dx }\]

\[I = \frac{1}{\sqrt{a}}\int\frac{1 - \sqrt{ax}}{1 - \sqrt{ax}} dx + \frac{1}{\sqrt{a}}\int\frac{a - 1}{1 - \sqrt{ax}} \text{ dx }\]

\[I = \frac{1}{\sqrt{a}}\int dx + \frac{a - 1}{\sqrt{a}}\int\frac{1}{1 - \sqrt{ax}} \text{ dx }\]

\[I = \frac{1}{\sqrt{a}}x + \frac{a - 1}{\sqrt{a}}\int\frac{1}{1 - \sqrt{ax}} \text{ dx}\]

\[\text{ Let,} \]

\[ I_1 = \int\frac{1}{1 - \sqrt{ax}} \text{ dx }\]

\[\text{ Put ax = z}^2 \]

\[ \Rightarrow adx = \text{ 2 }zdz\]

\[ I_1 = \frac{1}{a}\int\frac{2z}{1 - z}\text{  dz}\]

\[ I_1 = \frac{1}{a}\int\frac{2z - 2 + 2}{1 - z} \text{ dz }\]

\[ I_1 = \frac{1}{a}\int\frac{2z - 2}{1 - z} \text{ dz } + \frac{1}{a}\int\frac{2}{1 - z} \text{ dz }\]

\[ I_1 = \frac{- 2}{a}\int\frac{1 - z}{1 - z} \text{ dz } + \frac{1}{a}\int\frac{2}{1 - z} \text{ dz }\]

\[ I_1 = \frac{- 2}{a}\int \text{ dz } + \frac{1}{a}\int\frac{2}{1 - z} \text{ dz }\]

\[ I_1 = \frac{- 2}{a}z - \frac{2}{a}\text{ log }\left| 1 - z \right| + C_1 \]

\[ I_1 = \frac{- 2\sqrt{ax}}{a} - \frac{2}{a}\text{ log}\left| 1 - \sqrt{ax} \right| + C_1 \]

\[I = \frac{1}{\sqrt{a}}x + \frac{a - 1}{\sqrt{a}}\left( \frac{- 2\sqrt{ax}}{a} - \frac{2}{a}\text{ log }\left| 1 - \sqrt{ax} \right| \right) + C\]

 

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