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Let f(x)=\frac{x}{\left(1+x^n\right)^{1 / n}} for n \geq 2 and g(x)=\underbrace{(\text { fofo ... of })}_{\text {foctor } n \text { tims }}(x). Then \int x^{n-2} g(x) d x equals
 

Option: 1

\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+K


Option: 2

\frac{1}{n-1}\left(1+n x^{-3}\right)^{1-\frac{1}{n}}+K


Option: 3

\frac{1}{n(n+1)}\left(1+n x^n\right)^{1+\frac{1}{n}}+K


Option: 4

\frac{1}{n+1}\left(1+n x^n\right)^{1+\frac{1}{n}}+K


Answers (1)

Here f(x)=\frac{f(x)}{\left[1+f(x)^n\right]^{1 / a}}=\frac{x}{\left(1+2 x^n\right)^{[/ x}}

\begin{aligned} & f f(x)=\frac{x}{\left(1+3 x^n\right)^{1 / n}} \\ & \Rightarrow g(x)=(\text { fofo. of })(x)=\frac{x}{\text { anses }}\left(1+n x^n\right)^{1 / n} \end{aligned}

Hence I=\int x^{n-2} g(x) d x=\int \frac{x^{n-1} d x}{\left(1+n x^n\right)^{1 / n}}

\begin{aligned} & =\frac{1}{n^2} \int \frac{n^2 x^{n-1} d x}{\left(1+n x^n\right)^{1 / n}}=\frac{1}{n^2} \int \frac{\frac{d}{d x}\left(1+n x^n\right)}{\left(1+n x^n\right)^{1 / n}} d x \\ & \therefore 1=\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+k . \end{aligned}

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Sumit Saini

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