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∫ 1 √ 2 X + 3 + √ 2 X − 3 D X - Mathematics

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प्रश्न

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

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

Rationalise the denominator

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

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अध्याय 19: Indefinite Integrals - Exercise 19.03 [पृष्ठ २३]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.03 | Q 6 | पृष्ठ २३

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