The value of int frac{dx}{x(x^{n}+1)} is | vedclass

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(A) Let $I = \int \frac{dx}{x(x^{n}+1)}$.Multiply the numerator and denominator by $x^{n-1}$:$I = \int \frac{x^{n-1} dx}{x^{n}(x^{n}+1)}$.Let $t = x^{

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$\frac{1}{n} \log \left(\frac{x^{n}}{x^{n}+1}\right)+C$

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(A) Let $I = \int \frac{dx}{x(x^{n}+1)}$.Multiply the numerator and denominator by $x^{n-1}$:$I = \int \frac{x^{n-1} dx}{x^{n}(x^{n}+1)}$.Let $t = x^{n}$,then $dt = nx^{n-1} dx$,which implies $x^{n-1} dx = \frac{dt}{n}$.Substituting these into the integral:$I = \int \frac{dt/n}{t(t+1)} = \frac{1}{n} \int \frac{dt}{t(t+1)}$.Using partial fractions,$\frac{1}{t(t+1)} = \frac{1}{t} - \frac{1}{t+1}$.$I = \frac{1}{n} \int \left( \frac{1}{t} - \frac{1}{t+1} \right) dt$.$I = \frac{1}{n} (\log |t| - \log |t+1|) + C$.$I = \frac{1}{n} \log \left| \frac{t}{t+1} \right| + C$.Substituting $t = x^{n}$ back:$I = \frac{1}{n} \log \left( \frac{x^{n}}{x^{n}+1} \right) + C$.

The value of $\int \frac{dx}{x(x^{n}+1)}$ is

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