Question

If $\int \frac{dx}{x^2{(x^n+1)}^{\frac{(n-1)}{n}}}=-{\left[f\left(x\right)\right]}^{1/n}+c$ then $f\left(x\right)$ is

• A. $1+x^n$
• B. $1+x^{-n}$
• C. $x^n+x^{-n}$
• D. None of these

Answer 1

Answer: (B)

$1+x^{-n}$

We have,
$\int \frac{dx}{x^2{(x^n+1)}^{\frac{(n-1)}{n}}}=\int \frac{dx}{x^2\cdot x^{n-1}{(1+\frac{1}{x^n})}^{\frac{(n-1)}{n}}}$
$=\int \frac{dx}{x^{n+1}{(1+x^{-n})}^{\frac{(n-1)}{n}}}$
Put, $1+x^{-n}=t$
$\therefore -n{x}^{-n-1}dx=dt$
$\Rightarrow \frac{dx}{x^{n+1}}=\frac{-dt}{n}$
$\Rightarrow \int \frac{dx}{x^2{(x^n+1)}^{\frac{(n-1)}{n}}}=-\frac{1}{n}\int \frac{dt}{t^{\frac{(n-1)}{n}}}$
$=-\frac{1}{n}\int t^{\left(1/n\right)-1}dt=-\frac{1}{n}\cdot \frac{t^{(\frac{1}{n}-1+1)}}{(\frac{1}{n}-1+1)}+c$
$=-t^{1/n}+c$

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