Question

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

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

Answer 1

The correct option is B

$1+x^{-n}$

$\int \frac{dx}{x^2(x^n+1{)}^{\frac{(n-1)}{n}}}=\int \frac{dx}{x^2.x^{n-1}{(1+\frac{1}{x^n})}^{\frac{(n-1)}{n}}}$
$=\frac{dx}{x^{n+1}(1+x^{-n}{)}^{\frac{(n-1)}{n}}}$
Put $1+x^{-n}=t$
$\therefore -n{x}^{-n-1}dx=dt\text{or}\frac{dx}{x^{n+1}}=-\frac{dt}{n}$
$=\frac{dx}{x^{n+1}(1+x^{-n}{)}^{\frac{(n-1)}{n}}}=-\frac{1}{n}\int \frac{dt}{t^{\left(\frac{(n-1)}{n}\right)}}$
$=-\frac{1}{n}\int t^{\frac{1}{n}-1}dt=-\frac{1}{n}.\frac{t^{\frac{1}{n}-1+1}}{\frac{1}{n}-1+1}+c=-t^{\frac{1}{n}}$ +c
$=-(1+x^{-n}{)}^{\frac{1}{n}}$ +c