Question

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

Answer 1

\[\int\frac{dx}{\left( \sqrt{2x + 3} + \sqrt{2x} - 3 \right)}\]

Rationalise the denominator

\[= \int\frac{\left( \sqrt{2x + 3} - \sqrt{2x - 3} \right)}{\left( \sqrt{2x + 3} + \sqrt{2x - 3} \right)\left( \sqrt{2x + 3} - \sqrt{2x - 3} \right)}dx\]
\[ = \int\frac{\left( \sqrt{2x + 3} - \sqrt{2x - 3} \right)}{\left( 2x + 3 \right) - \left( 2x - 3 \right)}dx\]
\[ = \frac{1}{6}\int \left( 2x + 3 \right)^\frac{1}{2} dx - \frac{1}{6}\int \left( 2x - 3 \right)^\frac{1}{2} dx\]
\[ = \frac{1}{6}\left[ \frac{\left( 2x + 3 \right)^\frac{1}{2} + 1}{2\left( \frac{1}{2} + 1 \right)} \right] - \frac{1}{6}\left[ \frac{\left( 2x - 3 \right)^\frac{1}{2} + 1}{2\left( \frac{1}{2} + 1 \right)} \right] + C\]
\[ = \frac{1}{18}\left\{ \left( 2x + 3 \right)^\frac{3}{2} - \left( 2x - 3 \right)^\frac{3}{2} \right\} + C\]