If f(x) = frac{x}{(1 + nx^n)^{1/n}} for n geq | vedclass

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(A) Given $\int x^{n-2} f(x) dx = \int x^{n-2} \cdot \frac{x}{(1 + nx^n)^{1/n}} dx = \int \frac{x^{n-1}}{(1 + nx^n)^{1/n}} dx$. Let $t = 1 + nx^n$. Th

Vedclass · Accepted Answer

$\frac{1}{n(n-1)}(1 + nx^n)^{1 - 1/n} + C$

Vedclass · Answer

(A) Given $\int x^{n-2} f(x) dx = \int x^{n-2} \cdot \frac{x}{(1 + nx^n)^{1/n}} dx = \int \frac{x^{n-1}}{(1 + nx^n)^{1/n}} dx$. Let $t = 1 + nx^n$. Then $dt = n \cdot n x^{n-1} dx = n^2 x^{n-1} dx$,which implies $x^{n-1} dx = \frac{1}{n^2} dt$. Substituting these into the integral: $\int \frac{1}{n^2} \cdot t^{-1/n} dt = \frac{1}{n^2} \left[ \frac{t^{-1/n + 1}}{-1/n + 1} \right] + C$. $= \frac{1}{n^2} \left[ \frac{t^{(n-1)/n}}{(n-1)/n} \right] + C = \frac{1}{n^2} \cdot \frac{n}{n-1} \cdot t^{(n-1)/n} + C$. $= \frac{1}{n(n-1)} (1 + nx^n)^{1 - 1/n} + C$.

If $f(x) = \frac{x}{(1 + nx^n)^{1/n}}$ for $n \geq 2$,then $\int x^{n-2} f(x) dx =$

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