Advertisements
Advertisements
प्रश्न
उत्तर
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
संबंधित प्रश्न
` ∫ {sin 2x} /{a cos^2 x + b sin^2 x } ` dx
Evaluate the following integrals:
\[\int\frac{x + 2}{\left( x + 1 \right)^3} \text{ dx }\]
\[\int\frac{1}{2 - 3 \cos 2x} \text{ dx }\]
\[\int\frac{1}{\sin^4 x + \cos^4 x} \text{ dx}\]
\[\int\frac{1 + \sin x}{\sin x \left( 1 + \cos x \right)} \text{ dx }\]
\[\int\frac{x + 3}{\left( x + 4 \right)^2} e^x dx =\]
Find: `int (sin2x)/sqrt(9 - cos^4x) dx`
