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प्रश्न

\[\int\limits_0^1 \frac{1}{\sqrt{1 + x} - \sqrt{x}} dx\]

उत्तर

\[Let I = \int_0^1 \frac{1}{\sqrt{1 + x} - \sqrt{x}} d x . Then, \]
\[I = \int_0^1 \left( \frac{1}{\sqrt{1 + x} - \sqrt{x}} \times \frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} + \sqrt{x}} \right) d x\]
\[ \Rightarrow I = \int_0^1 \frac{\sqrt{1 + x} + \sqrt{x}}{1 + x - x} d x\]
\[ \Rightarrow I = \int_0^1 \left( \sqrt{1 + x} + \sqrt{x} \right) dx\]
\[ \Rightarrow I = \left[ \frac{2}{3} \left( 1 + x \right)^\frac{3}{2} + \frac{2}{3} x^\frac{3}{2} \right]_0^1 \]
\[ \Rightarrow I = \frac{2}{3} \times 2\sqrt{2} + \frac{2}{3} - \frac{2}{3}\]
\[ \Rightarrow I = \frac{4\sqrt{2}}{3}\]

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Definite Integrals
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अध्याय 20: Definite Integrals - Exercise 20.1 [पृष्ठ १७]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Exercise 20.1 | Q 54 | पृष्ठ १७

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