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प्रश्न
\[\int\frac{1}{\sqrt{x + a} + \sqrt{x + b}} dx\]
योग
उत्तर
\[\int\frac{dx}{\left( \sqrt{x + a} + \sqrt{x + b} \right)}\]
\[ = \int\frac{\left( \sqrt{x + a} - \sqrt{x - b} \right)}{\left( \sqrt{x + a} + \sqrt{x + b} \right)\left( \sqrt{x + a} - \sqrt{x + b} \right)}dx\]
\[ = \int\frac{\left( \sqrt{x + a} - \sqrt{x + b} \right)}{\left( x + a \right) - \left( x + b \right)}dx\]
\[ = \frac{1}{a - b}\int \left( x + a \right)^\frac{1}{2} - \frac{1}{a - b}\int \left( x + b \right)^\frac{1}{2} dx\]
` = 1/ ((a-b)) [ [(x+a ) ^ {1/2+1}] / [1/2 + 1]] ` - `1/(a-b) [[(x+b ) ^ {1/2+1}] / [1/2 + 1]] `
\[ = \frac{2}{3\left( a - b \right)}\left[ \left( x + a \right)^\frac{3}{2} - \left( x + b \right)^\frac{3}{2} \right] + C\]
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