Advertisements
Advertisements
Question
Advertisements
Solution
Rationalise the denominator
\[= \int\frac{\left( \sqrt{2x + 3} - \sqrt{2x - 3} \right)}{\left( \sqrt{2x + 3} + \sqrt{2x - 3} \right)\left( \sqrt{2x + 3} - \sqrt{2x - 3} \right)}dx\]
\[ = \int\frac{\left( \sqrt{2x + 3} - \sqrt{2x - 3} \right)}{\left( 2x + 3 \right) - \left( 2x - 3 \right)}dx\]
\[ = \frac{1}{6}\int \left( 2x + 3 \right)^\frac{1}{2} dx - \frac{1}{6}\int \left( 2x - 3 \right)^\frac{1}{2} dx\]
\[ = \frac{1}{6}\left[ \frac{\left( 2x + 3 \right)^\frac{1}{2} + 1}{2\left( \frac{1}{2} + 1 \right)} \right] - \frac{1}{6}\left[ \frac{\left( 2x - 3 \right)^\frac{1}{2} + 1}{2\left( \frac{1}{2} + 1 \right)} \right] + C\]
\[ = \frac{1}{18}\left\{ \left( 2x + 3 \right)^\frac{3}{2} - \left( 2x - 3 \right)^\frac{3}{2} \right\} + C\]
APPEARS IN
RELATED QUESTIONS
Integrate the following integrals:
\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
\[\int\frac{1}{\sqrt{x} + \sqrt{x + 1}} \text{ dx }\]
\[\int\frac{1 - x^4}{1 - x} \text{ dx }\]
