Mathematics

If $$\displaystyle \int { \frac { { x }^{ 2 }-x-1 }{ \sqrt [ 3 ]{ 1-3x }  } dx } =\frac { -\left( { x }^{ 2 }-x-1 \right) { \left( 1-3x \right)  }^{ \frac { 2 }{ 3 }  } }{ 2 } -\frac { \left( 2x-1 \right) { \left( 1-3x \right)  }^{ \frac { 8 }{ 3 }  } }{ k } +C$$, where $$k=$$


ANSWER

40


SOLUTION
$$\displaystyle I=\int { \frac { { x }^{ 2 }-x-1 }{ \sqrt [ 3 ]{ 1-3x }  } dx } =\int { \left( { x }^{ 2 }-x-1 \right) { \left( 1-3x \right)  }^{ \frac { -1 }{ 3 }  } } dx$$

$$\displaystyle=\left( { x }^{ 2 }-x-1 \right) \frac { { \left( 1-3x \right)  }^{ \frac { 2 }{ 3 }  } }{ \left( \frac { 2 }{ 3 }  \right) \left( -3 \right)  }-\int { \frac { { \left( 1-3x \right)  }^{ \frac { 2 }{ 3 }  } }{ \left( \frac { 2 }{ 3 }  \right) \left( -3 \right)  }  } .\left( 2x-1 \right) dx $$

$$\displaystyle=\frac { \left( { x }^{ 2 }-x-1 \right) { \left( 1-3x \right)  }^{ \frac { 2 }{ 3 }  } }{ -2 } +\frac { 1 }{ 2 } \int { \left( 2x-1 \right)  } { \left( 1-3x \right)  }^{ \frac { 2 }{ 3 }  }dx$$

$$\displaystyle=\frac { \left( { x }^{ 2 }-x-1 \right) { \left( 1-3x \right)  }^{ \frac { 2 }{ 3 }  } }{ -2 } +\frac { 1 }{ 2 } \left( 2x-1 \right) \frac { { \left( 1-3x \right)  }^{ \frac { 5 }{ 3 }  } }{ \left( \frac { 5 }{ 3 }  \right) \left( -3 \right)  }-\frac { 1 }{ 2 } \int { \frac { { \left( 1-3x \right)  }^{ \frac { 5 }{ 3 }  } }{ \left( \frac { 5 }{ 3 }  \right) \left( -3 \right)  }  }  $$

$$\displaystyle=\frac { \left( { x }^{ 2 }-x-1 \right) { \left( 1-3x \right)  }^{ \frac { 2 }{ 3 }  } }{ -2 } +\frac { \left( 2x-1 \right) { \left( 1-3x \right)  }^{ \frac { 5 }{ 3 }  } }{ -10 } +\frac { 1 }{ 5 } \frac { { \left( 1-3x \right)  }^{ \frac { 8 }{ 3 }  } }{ \left( \frac { 8 }{ 3 }  \right) \left( -3 \right)  } +C$$

$$\displaystyle=\frac { \left( { x }^{ 2 }-x-1 \right) { \left( 1-3x \right)  }^{ \frac { 2 }{ 3 }  } }{ -2 } +\frac { \left( 2x-1 \right) { \left( 1-3x \right)  }^{ \frac { 5 }{ 3 }  } }{ -10 } +\frac { { \left( 1-3x \right)  }^{ \frac { 8 }{ 3 }  } }{ -40 } +C$$
$$\therefore k=40$$
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