If $f(x)=\frac{x}{\left(1+n x^n\right)^{1 / n}}$ for $n \geq 2$, then $\int x^{n-2} f(x) d x=$
If $f(x)=\frac{x}{\left(1+n x^n\right)^{1 / n}}$ for $n \geq 2$, then $\int x^{n-2} f(x) d x=$
- $\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+C$
- $\frac{1}{(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+C$
- $\frac{1}{n(n-1)}\left(1+n x^n\right)^{1+\frac{1}{n}}+C$
- $\frac{1}{n+1}\left(1+n x^n\right)^{1+\frac{1}{n}}+C$
Solution
$\int x^{n-2} f(x) d x=\int \frac{x^{n-1}}{\left(1+n x^n\right)^{1 / n}} d x$
Let $1+n x n=\mathrm{t} \Rightarrow x^{n-1} d x=\frac{1}{n^2} d t$
$
\begin{aligned}
& =\frac{1}{n^2} \int \frac{d t}{t^{1 / n}}=\frac{1}{n^2} \frac{t^{-1 / n}+1}{\left(-\frac{1}{n}+1\right)}+C \\
& =\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+C
\end{aligned}
$
Asked in: AP EAMCET 2023 (19 May Shift 1)
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