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