Question

\[\int\frac{1}{\sqrt{x} + \sqrt{x + 1}}  \text{ dx }\]

Answer 1

\[\int\frac{1}{\sqrt{x + 1} + \sqrt{x}}dx\]
\[\text{Rationalising the denominator}, \]
\[ = \int\frac{\left( \sqrt{x + 1} - \sqrt{x} \right)}{\left( \sqrt{x + 1} + \sqrt{x} \right) \left( \sqrt{x + 1} - \sqrt{x} \right)}dx\]
\[ = \int\frac{\left( \sqrt{x + 1} - \sqrt{x} \right)}{x + 1 - x}dx\]
\[ = \int\left( \sqrt{x + 1} - \sqrt{x} \right) dx\]
\[ = \int\left[ \left( x + 1 \right)^\frac{1}{2} - x^\frac{1}{2} \right]dx\]
\[ = \frac{\left( x + 1 \right)^\frac{1}{2} + 1}{\frac{1}{2} + 1} - \frac{x^\frac{1}{2} + 1}{\frac{1}{2} + 1} + C\]
\[ = \frac{2}{3} \left( x + 1 \right)^\frac{3}{2} - \frac{2}{3} x^\frac{3}{2} + C\]
\[ = \frac{2}{3}\left[ \left( x + 1 \right)^\frac{3}{2} - x^\frac{3}{2} \right] + C\]