Q117 / 200 · Manipal (MET) 2013 · Mathematics
If $\int \frac{\text { Th }}{x^2\left(x^n+1\right)^{\frac{1-1)}{n}}}:=-[f(x)]^{1 / n}+c$ then $f(x)$ is
- A$1+x^n$
- Correct answerB$1+x^{-n}$
- C$x^n+x^{-n}$
- DNone of these
Solution
We have;
$\int \frac{d x}{x^2\left(x^n+1\right)^{\frac{(n-1)}{n}}}=\int \frac{d x}{x^2 \cdot x^{n-1}\left(1+\frac{1}{x^n}\right)^{\frac{(n-1)}{n}}}$
$=\int \frac{d x}{x^{n+1}\left(1+x^{-n}\right)^{\frac{(n-1)}{n}}}$
Put, $1+x^{-n}=t$
$\therefore \quad-n x^{-n-1} d x=d t$
$\Rightarrow \quad \frac{d x}{x^{n+1}}=\frac{-d t}{n}$
$\Rightarrow \quad \int \frac{d x}{x^2\left(x^n+1\right)^{\frac{(n-1)}{n}}}=-\frac{1}{n} \int \frac{d t}{t^{\frac{(n-1)}{n}}}$
$=-\frac{1}{n} \int t^{(1 / n)-1} d t=-\frac{1}{n} \cdot \frac{t^{\left(\frac{1}{n}-1+1\right)}}{\left(\frac{1}{n}-1+1\right)}+c$
$=-t^{1 / n}+c=-\left(1+x^{-n}\right)^{\frac{1}{n}}+c$
Hence, the correct answer is option: 'B'.