Question

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

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

Answer 1

Answer: (A)

$\frac{1}{n(n-1)}(1+n{x}^n{)}^{1-\frac{1}{n}}+c$

$\begin{array}{c}\Rightarrow f\left(x\right)=\frac{x}{{(1+x^n)}^{1/n}},n⩾2\\ g\left(x\right)=f\circ f0\dots \text{of}\left(x\right)\\ \therefore f\circ f\left(x\right)=f\left(\frac{x}{{(1+x^n)}^{1/n}}\right)=\frac{{(1+x^n)}^{1/n}}{[1+{\left(\frac{x}{{(1+x^n)}^{1/n})}\right]}^{n/n}}\\ =\frac{x}{{(1+2x^n)}^{1/n}}\end{array}$

similarly,

$\text{fofo.}\text{of}\left(x\right)=\frac{x}{{(1+n{x}^n)}^{1/n}}=g\left(x\right)$

Now, $\begin{array}{c}\Rightarrow \int x^{n-2}g\left(x\right)\cdot dx\\ =\int x^{n-2}\cdot \frac{x}{{(1+n{x}^n)}^{1/n}}\cdot dx=\int \frac{x^{n-1}}{(1+nx{)}^{1/n}}\cdot dx\\ \begin{array}{c}{n}^2{x}^{n-1}dx=n{t}^{n-1}dt\\ n{x}^{n-1}dx=t^{n-1}dt\end{array}\end{array}$

$\begin{array}{c}=\int \frac{t^{n-1}}{nt}dt=\frac{1}{n}\int t^{n-2}dt=\frac{1}{n}\left(\frac{t^{n-1}}{n-1}\right)+c\\ \text{Substitute value of}t\text{we get}\\ \int x^{n-2}g\left(x\right)\cdot dx=\frac{1}{n(n-1)}{(1+n{x}^n)}^{1-\frac{1}{n}}+c\\ \text{Hence,}\left(A\right)\text{is the correct option.}\end{array}$