हिंदी

∫ 1 √ X + 3 − √ X + 2 D X

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प्रश्न

\[\int\frac{1}{\sqrt{x + 3} - \sqrt{x + 2}} dx\]
योग
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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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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 18: Indefinite Integrals - Exercise 19.03 [पृष्ठ २३]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 18 Indefinite Integrals
Exercise 19.03 | Q 17 | पृष्ठ २३
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