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प्रश्न
` ∫ 1 /{x^{1/3} ( x^{1/3} -1)} ` dx
बेरीज
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उत्तर
` Let I = ∫ 1 /{x^{1/3} ( x^{1/3} -1)} ` dx
` ∫ 1 / {x^{2/3} - x^{1/3}} ` dx
\[ \text{Let x }= t^3 \]
\[ \text{On differentiating both sides, we get}\]
\[ dx = 3 t^2 \text{ dt }\]
` ∴ I = ƒ {3 t^2}/ {( t^3 )^\{2/3} - ( t^3)^{1/3}} dt `
\[ = \int\frac{3 t^2}{t^2 - t}dt\]
\[ = 3\int\frac{t}{t - 1}dt\]
\[ = 3\int\frac{\left( t - 1 \right) + 1}{t - 1}dt\]
\[ = 3\int\left[ \left( 1 \right) + \frac{1}{t - 1} \right]dt\]
\[ = 3\left[ t + \text{ log }\left( t - 1 \right) \right] + c\]
` = 3 x ^{1/3 } + 3 log ( x^{1/3} -1 ) + c `
Hence, ` ∫ 1 /{x^{1/3} ( x^{1/3} -1)} dx = 3x^{1/3} + 3 log ( x^{1/3} -1 ) + c `
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