Question

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

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

Answer 1

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

$\begin{array}{c}f\left(x\right)=\frac{x}{{\left(1+x^n\right)}^{\frac{1}{n}}}\\ fof\left(x\right)=\frac{\frac{x}{{\left(1+x^n\right)}^{\frac{1}{n}}}}{{\left(1+\frac{x^n}{1+x^n}\right)}^{\frac{1}{n}}}=\frac{x}{\left(2+x^n\right)\frac{1}{n}}\\ \underset{3times}{\underset{}{fofof}}\left(x\right)=\frac{\frac{x}{{\left(1+x^n\right)}^{\frac{1}{n}}}}{{\left(2+\frac{x^n}{1+x^n}\right)}^{\frac{1}{n}}}=\frac{x}{{\left(3+x^n\right)}^{\frac{1}{n}}}\\ g\left(x\right)=\underset{ntimes}{\underset{}{fofof.....f}}\left(x\right)=\frac{x}{{\left(n+x^n\right)}^{\frac{1}{n}}}\\ \int x^{n-2}g\left(x\right)dx=\int \frac{x^{n-1}}{{\left(n+x^n\right)}^{\frac{1}{n}}}dx\\ \left[\begin{array}{c}n+x^n=t^n\\ \Rightarrow x^{n-1}dx=t^{n-1}dt\end{array}\right]\\ =\int \frac{t^{n-1}dt}{{\left(t^n\right)}^{\frac{1}{n}}}=\int t^{n-2}dt\\ =\frac{t^{n-1}}{n-1}\\ =\frac{1}{n-1}{\left(n+x^n\right)}^{\frac{n+1}{n}}+C\end{array}$