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प्रश्न
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उत्तर
\[\text{ Let I } = \int\sqrt{\frac{a + x}{x}}dx\]
\[ = \int\frac{\sqrt{\left( a + x \right) \left( a + x \right)}}{\sqrt{x \left( a + x \right)}}\]
\[ = \int\left( \frac{a + x}{\sqrt{x^2 + ax}} \right)dx\]
\[ = a\int\frac{1}{\sqrt{x^2 + ax}}\text{ dx} + \int\frac{x}{\sqrt{x^2 + ax}}\text{ dx}\]
\[ = a\int\frac{1}{\sqrt{x^2 + ax + \left( \frac{a}{2} \right)^2 - \left( \frac{a}{2} \right)^2}}\text{ dx} + \int\frac{x}{\sqrt{x^2 + ax}}\text{ dx}\]
\[ = a\int\frac{1}{\sqrt{\left( x + \frac{a}{2} \right)^2 - \left( \frac{a}{2} \right)^2}}\text{ dx}+ \frac{1}{2}\int\frac{2x}{\sqrt{x^2 + ax}}\text{ dx}\]
\[ = a\int\frac{1}{\sqrt{\left( x + \frac{a}{2} \right)^2 - \left( \frac{a}{2} \right)^2}}\text{ dx}+ \frac{1}{2}\int\left( \frac{2x + a - a}{\sqrt{x^2 + ax}} \right)\text{ dx}\]
\[ = a\int\frac{1}{\sqrt{\left( x + \frac{a}{2} \right)^2 - \left( \frac{a}{2} \right)^2}}\text{ dx } + \frac{1}{2}\int\frac{\left( 2x + a \right)}{\sqrt{x^2 + ax}}\text{ dx }- \frac{a}{2}\int\frac{1}{\sqrt{x^2 + ax}}\text{ dx }\]
\[ = \frac{a}{2}\int\frac{1}{\sqrt{\left( x + \frac{a}{2} \right)^2 - \left( \frac{a}{2} \right)^2}}\text{ dx } + \frac{1}{2}\int\frac{\left( 2x + a \right)}{\sqrt{x^2 + ax}} \text{ dx }\]
\[\text{ Putting x}^2 + ax = \text{ t in the Ist integral} \]
\[ \Rightarrow \left( 2x + a \right) dx = dt\]
\[ \therefore I = \frac{a}{2}\int\frac{1}{\sqrt{\left( x + \frac{a}{2} \right)^2 - \left( \frac{a}{2} \right)^2}}dx + \frac{1}{2}\int\frac{1}{\sqrt{t}}dt\]
\[ = \frac{a}{2} \text{ ln }\left| x + \frac{a}{2} + \sqrt{x^2 + ax} \right| + \frac{1}{2} \times 2\sqrt{t} + C .................\left[ \because \int\frac{1}{\sqrt{x^2 - a^2}}dx = \text{ ln}\left| x + \sqrt{x^2 - a^2} \right| + C \right]\]
\[ = \frac{a}{2} \text{ ln } \left| x + \frac{a}{2} + \sqrt{x^2 + ax} \right| + \sqrt{x^2 + ax} + C ..........\left[ \because t = x^2 + ax \right]\]
