int frac{dx}{x(x^n + 1)} = | vedclass

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(C) 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 $x^n = t$,

Vedclass · Accepted Answer

$\frac{1}{n} \log \frac{x^n}{x^n + 1} + c$

Vedclass · Answer

(C) 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 $x^n = t$,then $n x^{n-1} dx = dt$,which implies $x^{n-1} dx = \frac{dt}{n}$.Substituting these into the integral:$I = \frac{1}{n} \int \frac{dt}{t(t+1)}$.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:$I = \frac{1}{n} \log \frac{x^n}{x^n + 1} + c$.

$\int \frac{dx}{x(x^n + 1)} = $

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