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 $\displaystyle \int x^{n-2}g\left ( x \right )dx$

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

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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