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Question
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Solution
\[\text{Let I }=\int_\frac{1}{3}^1 \frac{\left( x - x^3 \right)^\frac{1}{3}}{x^4}dx\]
\[= \int_\frac{1}{3}^1 \frac{\left[ x^3 \left( \frac{x}{x^3} - 1 \right) \right]^\frac{1}{3}}{x^4}dx\]
\[ = \int_\frac{1}{3}^1 \frac{x \left( \frac{1}{x^2} - 1 \right)^\frac{1}{3}}{x^4}dx\]
\[ = \int_\frac{1}{3}^1 \frac{\left( \frac{1}{x^2} - 1 \right)^\frac{1}{3}}{x^3}dx\]
Put
\[\therefore - \frac{2}{x^3}dx = dz\]
\[ \Rightarrow \frac{dx}{x^3} = - \frac{dz}{2}\]
When
When
\[\therefore I = - \frac{1}{2} \int_8^0 z^\frac{1}{3} dz\]
\[ = \left.- \frac{1}{2} \times \frac{z^\frac{4}{3}}{\frac{4}{3}}\right|_8^0 \]
\[ = - \frac{3}{8}\left[ 0 - \left( 8 \right)^\frac{4}{3} \right]\]
\[ = - \frac{3}{8} \times \left( - 16 \right)\]
\[ = 6\]
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