int frac{dx}{x(x^5 + 1)} = | vedclass

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Question

(C) Let $I = \int \frac{dx}{x(x^5 + 1)}$.Multiply the numerator and denominator by $x^4$:$I = \int \frac{x^4 dx}{x^5(x^5 + 1)}$.Let $x^5 = t$,then $5x

Vedclass · Accepted Answer

$\frac{1}{5}\log |\frac{x^5}{x^5 + 1}| + c$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{x(x^5 + 1)}$.Multiply the numerator and denominator by $x^4$:$I = \int \frac{x^4 dx}{x^5(x^5 + 1)}$.Let $x^5 = t$,then $5x^4 dx = dt$,which implies $x^4 dx = \frac{dt}{5}$.Substituting these into the integral:$I = \frac{1}{5} \int \frac{dt}{t(t + 1)}$.Using partial fractions:$\frac{1}{t(t + 1)} = \frac{1}{t} - \frac{1}{t + 1}$.Therefore,$I = \frac{1}{5} \int (\frac{1}{t} - \frac{1}{t + 1}) dt$.$I = \frac{1}{5} (\log |t| - \log |t + 1|) + c$.$I = \frac{1}{5} \log |\frac{t}{t + 1}| + c$.Substituting $t = x^5$ back:$I = \frac{1}{5} \log |\frac{x^5}{x^5 + 1}| + c$.Comparing this with the given options,option $C$ is correct.

$\int \frac{dx}{x(x^5 + 1)} = $

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