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