Question

Let $f\left(x\right)=\frac{x}{{(1+x^n)}^{1/n}}$ for $n\ge 2$ and $g\left(x\right)=\underset{foccursntimes}{\underset{}{(f_o{f}_o...f_o)}}\left(x\right)$. Then $\int x^{n-2}g\left(x\right)dx$ equals?

• A. $\frac{1}{n(n-1)}{(1+n{x}^n)}^{1-\frac{1}{n}}+K$
• B. $\frac{1}{(n-1)}{(1+n{x}^n)}^{1-\frac{1}{n}}+K$
• C. $\frac{1}{n(n+1)}{(1+n{x}^n)}^{1+\frac{1}{n}}+K$
• D. $\frac{1}{(n+1)}{(1+n{x}^n)}^{1+\frac{1}{n}}+K$

Answer 1

The correct option is A $\frac{1}{n(n-1)}{(1+n{x}^n)}^{1-\frac{1}{n}}+K$

$f\left(x\right)=\frac{x}{(1+x^n{)}^{\frac{1}{n}}}$

$f\left(f\right(x\left)\right)=\frac{\frac{x}{(1+x^n{)}^{\frac{1}{n}}}}{{(1+\frac{x^n}{1+x^n})}^{\frac{1}{n}}}$

$f\left(f\right(x\left)\right)=\frac{x}{(1+2x^n{)}^{\frac{1}{n}}}$

Similarly

$f\left(f\right(x\left)\right)=\frac{x}{(1+n{x}^n{)}^{\frac{1}{n}}}$

$=\int \frac{x^{n-2}\times x}{(1+n{x}^n{)}^{\frac{1}{n}}}=\int \frac{x^{n-1}dx}{(1+n{x}^n{)}^{\frac{1}{n}}}$

$1+x^n=t$

$n{x}^{n-1}dx=dt$

$=\frac{1}{n}\int \frac{dt}{\left(t\right)\frac{1}{n}}$

$\frac{1}{n}\frac{(t{)}^{1-\frac{1}{n}}}{(1-\frac{1}{n})}=\frac{n}{n(n-1)}(1+n{2}^n{)}^{1-\frac{1}{n}}$.