Mathematics

The value of the integral $$\displaystyle\int^{1}_{\dfrac{1}{3}}\dfrac{(x-x^3)^{\dfrac{1}{3}}}{x^4}dx$$ is?


ANSWER

$$6$$


SOLUTION
\begin{array}{l} \int  _{ \frac { 1 }{ 3 }  }^{ 1 }{ \frac { { { { (x-{ x^{ 3 } }) }^{ \frac { 1 }{ 3 }  } } } }{ { { x^{ 4 } } } }  }dx \\ =\int  _{ \frac { 1 }{ 3 }  }^{ 1 }{ \frac { { { x^{ 3 } }\times \frac { 1 }{ 3 } { { (\frac { 1 }{ { { x^{ 2 } } } } -1) }^{ \frac { 1 }{ 3 }  } } } }{ { { x^{ 4 } } } }  }dx \\ =\int  _{ \frac { 1 }{ 3 }  }^{ 1 }{ x\frac { { { { \left( { \frac { 1 }{ { { x^{ 2 } } } } -1 } \right)  }^{ \frac { 1 }{ 3 }  } } } }{ { { x^{ 4 } } } }  }dx \\ =\int  _{ \frac { 1 }{ 3 }  }^{ 1 }{ \frac { 1 }{ { { x^{ 3 } } } } \left( { \frac { 1 }{ { { x^{ 2 } } } } -1 } \right)  }dx \\ let{ \left( { \frac { 1 }{ { { x^{ 2 } }-1 } }  } \right) ^{ \frac { 1 }{ 3 }  } }=t \\ \frac { 1 }{ 3 } { \left( { \frac { 1 }{ { { x^{ 2 } } } } -1 } \right) ^{ \frac { { -2 } }{ 3 }  } }\cdot \left( { -\frac { 1 }{ { { x^{ 3 } } } }  } \right) dx=d \\ =-3\int  _{ \frac { 1 }{ 3 }  }^{ 1 }{ \frac { 1 }{ { { x^{ 3 } } } }  }t\cdot { x^{ 3 } }{ \left( { \frac { 1 }{ { { x^{ 2 } } } } -1 } \right) ^{ \frac { 2 }{ 3 }  } }dt \\ =-3\int  _{ \frac { 1 }{ 3 }  }^{ 1 }{ t\cdot { t^{ 2 } }dt } \\ =-3\int  _{ \frac { 1 }{ 3 }  }^{ 1 }{ { t^{ 3 } }dt } \\ =-3\left[ { \frac { { { t^{ 4 } } } }{ 4 }  } \right] _{ \frac { 1 }{ 3 }  }^{ 1 } \\ =-3\left[ { { { \left( { \frac { 1 }{ { { x^{ 2 } }-1 } }  } \right)  }^{ \frac { 1 }{ 3 }  } } } \right] _{ \frac { 1 }{ 3 }  }^{ 1 } \\ =3\left[ { 0-2 } \right]  \\ =6 \end{array}
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Single Correct Medium Published on 17th 09, 2020
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