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प्रश्न
\[\int\frac{1}{\sqrt{x + 1} + \sqrt{x}} dx\]
योग
उत्तर
\[\int\frac{dx}{\sqrt{x + 1} + \sqrt{x}}\]
Rationalise the denominator
\[= \int\frac{\left( \sqrt{x + 1} - \sqrt{x} \right)}{\left( \sqrt{x + 1} + \sqrt{x} \right)\left( \sqrt{x + 1} - \sqrt{x} \right)}dx\]
\[ = \int\frac{\left( \sqrt{x + 1} - \sqrt{x} \right)}{\left( x + 1 \right) - x}dx\]
\[ = \int \left( x + 1 \right)^\frac{1}{2} dx - \int x^\frac{1}{2} dx\]
\[ = \frac{\left( x + 1 \right)^\frac{1}{2} + 1}{\frac{1}{2} + 1} - \frac{x^\frac{1}{2} + 1}{\frac{1}{2} + 1}\]
\[ = \frac{2}{3} \left( x + 1 \right)^\frac{3}{2} - \frac{2}{3} x^\frac{3}{2} + C\]
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