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प्रश्न
\[\int\frac{x^{- 1/3} + \sqrt{x} + 2}{\sqrt[3]{x}} dx\]
योग
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उत्तर
\[\int \left( \frac{x^{- \frac{1}{3}} + \sqrt{x} + 2}{x^\frac{1}{3}} \right)dx\]
\[ = \int \left( \frac{x^{- \frac{1}{3}}}{x^\frac{1}{3}} + \frac{x^\frac{1}{2}}{x^\frac{1}{3}} + \frac{2}{x^\frac{1}{3}} \right)dx\]
\[ = \int\left( x^{- \frac{2}{3}} + x^\frac{1}{6} + 2 x^{- \frac{1}{3}} \right)dx\]
\[ = \left[ \frac{x^{- \frac{2}{3} + 1}}{- \frac{2}{3} + 1} + \frac{x^\frac{1}{6} + 1}{\frac{1}{6} + 1} + 2\frac{x^{- \frac{1}{3} + 1}}{- \frac{1}{3} + 1} \right]\]
\[ = \left[ \frac{x^\frac{1}{3}}{\frac{1}{3}} + \frac{x^\frac{7}{6}}{\frac{7}{6}} + 3 x^\frac{2}{3} \right] + C\]
\[ = 3 x^\frac{1}{3} + \frac{6}{7} x^\frac{7}{6} + 3 x^\frac{2}{3} + C\]
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