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

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Question

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

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

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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Chapter 19: Indefinite Integrals - Exercise 19.03 [Page 23]

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RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.03 | Q 17 | Page 23

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