Let $f(x) = \frac{x}{\left( 1 + x^{n} \right)^{\frac{1}{n}}}$ for $n \geq 2$ and $g(x) = (fofo\ldots of)(x)$. [image:7_image1.emf] Then, $\int_{}^{}{x^{n - 2}\ g(x)dx}$ equals

Question

Dr. Abubacker · Accepted Answer

(a)
Here, $ff(x) = \frac{f(x)}{\left\lbrack 1 + f(x)^{n} ightbrack^{\frac{1}{n}}} = \frac{x}{\left( 1 + 2x^{n} ight)^{\frac{1}{n}}}$
and $fff(x) = \frac{x}{\left( 1 + 3x^{n} ight)^{\frac{1}{n}}}$
$	herefore\  = \frac{x}{\left( 1 + nx^{n} ight)^{\frac{1}{n}}}$
Let $I = \int_{}^{}{x^{n - 2}\ g(x)dx} = \int_{}^{}\frac{x^{n - 1}dx}{\left( 1 + nx^{n} ight)^{\frac{1}{n}}}$
$= \frac{1}{n^{2}}\int_{}^{}\frac{n^{2}x^{n - 1}dx}{\left( 1 + nx^{n} ight)^{\frac{1}{n}}}$
$= \frac{1}{n^{2}}\int_{}^{}\frac{\frac{d}{dx}(1 + nx^{n})}{\left( 1 + nx^{n} ight)^{\frac{1}{n}}}\ dx$
$	herefore\ I = \frac{1}{n(n - 1)}\left( 1 + nx^{n} ight)^{1 - \frac{1}{n}} + c$

Q. Let $f(x) = \frac{x}{\left( 1 + x^{n} \right)^{\frac{1}{n}}}$ for $n \geq 2$ and $g(x) = (fofo\ldots of)(x)$.

Then, $\int_{}^{}{x^{n - 2}\ g(x)dx}$ equals

Explanation

Reviewed by Dr. Abubacker

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Q. Let $f(x) = \frac{x}{\left( 1 + x^{n} \right)^{\frac{1}{n}}}$ for $n \geq 2$ and $g(x) = (fofo\ldots of)(x)$. Then, $\int_{}^{}{x^{n - 2}\ g(x)dx}$ equals

Explanation

Reviewed by Dr. Abubacker

Want more practice?

Q. Let $f(x) = \frac{x}{\left( 1 + x^{n} \right)^{\frac{1}{n}}}$ for $n \geq 2$ and $g(x) = (fofo\ldots of)(x)$.

Then, $\int_{}^{}{x^{n - 2}\ g(x)dx}$ equals