int frac{1}{x^m sqrt[m]{x^m+1}} d x= | vedclass

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(B) To solve the integral $I = \int \frac{1}{x^m (x^m+1)^{1/m}} dx$, factor out $x^m$ from the radical: $I = \int \frac{1}{x^m \cdot x (1 + x^{-m})^{1

Vedclass · Accepted Answer

$\frac{-1}{m-1}\left(\frac{\sqrt[m]{x^m+1}}{x}\right)^{m-1}+c$

Vedclass · Answer

(B) To solve the integral $I = \int \frac{1}{x^m (x^m+1)^{1/m}} dx$, factor out $x^m$ from the radical: $I = \int \frac{1}{x^m \cdot x (1 + x^{-m})^{1/m}} dx = \int \frac{1}{x^{m+1} (1 + x^{-m})^{1/m}} dx$. Let $u = 1 + x^{-m}$. Then $du = -m x^{-m-1} dx = -m x^{-(m+1)} dx$. Thus, $dx / x^{m+1} = -du / m$. Substituting these into the integral: $I = \int \frac{1}{u^{1/m}} \cdot \left(-\frac{du}{m}\right) = -\frac{1}{m} \int u^{-1/m} du$. Integrating with respect to $u$: $I = -\frac{1}{m} \cdot \frac{u^{1 - 1/m}}{1 - 1/m} + C = -\frac{1}{m} \cdot \frac{u^{(m-1)/m}}{(m-1)/m} + C = -\frac{1}{m-1} u^{(m-1)/m} + C$. Substituting $u = 1 + x^{-m} = \frac{x^m+1}{x^m}$ back: $I = -\frac{1}{m-1} \left(\frac{x^m+1}{x^m}\right)^{(m-1)/m} + C = -\frac{1}{m-1} \left(\frac{(x^m+1)^{1/m}}{x}\right)^{m-1} + C$. This matches option $B$.

$\int \frac{1}{x^m \sqrt[m]{x^m+1}} d x=$

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