Mathematics

 Let $$\displaystyle f\left ( x \right )= \frac{x}{(1+x^{n})}^{1/n}$$ for $$\displaystyle n \geq 2$$ and $$\displaystyle g\left ( x \right )= \underset{f occurs n times}{\underbrace{\left ( f\circ f\circ ...\circ f  \right )\left ( x \right )}}.$$ Then $$\displaystyle \int x^{n-2}g\left ( x \right )dx$$


ANSWER

$$\displaystyle \frac{1}{n\left ( n-1 \right )}\left ( 1+nx^{n} \right )^{1-\frac{1}{n}}+k$$


SOLUTION
Given $$\displaystyle f\left( x \right) =\frac { x }{ { \left( 1+x \right)  }^{ \frac { 1 }{ n }  } } $$ for $$n\ge 2$$

$$\displaystyle \therefore ff\left( x \right) =\frac { f\left( x \right)  }{ { \left[ 1+{ f\left( x \right)  }^{ n } \right]  }^{ \frac { 1 }{ n }  } } =\frac { x }{ { \left( 1+2x \right)  }^{ \frac { 1 }{ n }  } } $$ 

and $$\displaystyle fff\left( x \right) =\frac { x }{ { \left( 1+3{ x }^{ n } \right)  }^{ \frac { 1 }{ n }  } } $$

$$\displaystyle \therefore g\left( x \right) =\left( fofo...of \right) \left( x \right) =\frac { x }{ { \left( { 1+nx }^{ n } \right)  }^{ \frac { 1 }{ n }  } } $$

Let $$\displaystyle I=\int { { x }^{ n-2 } } g\left( x \right) dx=\int { \frac { { x }^{ n-1 }dx }{ { \left( 1+{ nx }^{ n } \right)  }^{ \frac { 1 }{ n }  } }  } $$

$$\displaystyle =\frac { 1 }{ { n }^{ 2 } } \int { \frac { { n }^{ 2 }{ x }^{ n-1 }dx }{ { \left( { 1+nx }^{ n } \right)  }^{ \frac { 1 }{ n }  } }  } =\frac { 1 }{ { n }^{ 2 } } \int { \frac { \dfrac { d }{ dx } \left( 1+{ nx }^{ n } \right)  }{ { \left( { 1+nx }^{ n } \right)  }^{ \frac { 1 }{ n }  } }  } dx$$

$$\displaystyle \therefore I=\frac { 1 }{ n\left( n-1 \right)  } { \left( { 1+nx }^{ n } \right)  }^{ 1-\frac { 1 }{ n }  }+c$$
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Single Correct Hard Published on 17th 09, 2020
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