int frac{dx}{x(x^7 + 1)} = | vedclass

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(B) Given integral is $I = \int \frac{dx}{x(x^7 + 1)}$.Multiply the numerator and denominator by $x^6$ to get $I = \int \frac{x^6 dx}{x^7(x^7 + 1)}$.L

Vedclass · Accepted Answer

$\frac{1}{7}\log \left( \frac{x^7}{x^7 + 1} \right) + c$

Vedclass · Answer

(B) Given integral is $I = \int \frac{dx}{x(x^7 + 1)}$.Multiply the numerator and denominator by $x^6$ to get $I = \int \frac{x^6 dx}{x^7(x^7 + 1)}$.Let $x^7 = u$,then $7x^6 dx = du$,which implies $x^6 dx = \frac{du}{7}$.Substituting these into the integral,we get $I = \int \frac{du/7}{u(u + 1)} = \frac{1}{7} \int \frac{du}{u(u + 1)}$.Using partial fractions,$\frac{1}{u(u + 1)} = \frac{1}{u} - \frac{1}{u + 1}$.Thus,$I = \frac{1}{7} \int \left( \frac{1}{u} - \frac{1}{u + 1} \right) du = \frac{1}{7} (\log |u| - \log |u + 1|) + c$.$I = \frac{1}{7} \log \left| \frac{u}{u + 1} \right| + c$.Substituting $u = x^7$ back,we get $I = \frac{1}{7} \log \left( \frac{x^7}{x^7 + 1} \right) + c$.

$\int \frac{dx}{x(x^7 + 1)} = $

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