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प्रश्न

\[\int\frac{1}{x^{2/3} \sqrt{x^{2/3} - 4}} dx\]
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उत्तर

 

\[ = \int\frac{dx}{x^\frac{2}{3} \sqrt{\left( x^\frac{1}{3} \right)^2 - 2^2}}\]

\[ = \int\frac{dx}{x^\frac{2}{3} \sqrt{\left( x^\frac{1}{3} \right)^2 - 2^2}}\]
\[\text{ Let } x^\frac{1}{3} = t\]
\[ \Rightarrow \frac{1}{3} x^\frac{- 2}{3} dx = dt\]
\[ \Rightarrow \frac{1}{3 x^\frac{2}{3}} dx = dt\]
\[ \Rightarrow \frac{dx}{x^\frac{2}{3}} = 3 dt\]
\[Now, \int\frac{dx}{x^\frac{2}{3} \sqrt{x^\frac{2}{3} - 2^2}}\]
\[ = 3\int\frac{dt}{\sqrt{t^2 - 2^2}}\]
\[ = 3 \text{ log } \left| t + \sqrt{t^2 - 2^2} \right| + C\]
\[ = 3 \text{ log }\left| x^\frac{1}{3} + \sqrt{x^\frac{2}{3} - 4} \right| + C\]

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अध्याय 19: Indefinite Integrals - Exercise 19.18 [पृष्ठ ९९]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.18 | Q 13 | पृष्ठ ९९

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