int frac{dx}{x(x^4 - 1)} is equal to :- | vedclass

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(A) Let $I = \int \frac{dx}{x(x^4 - 1)}$.Multiply the numerator and denominator by $x^3$:$I = \int \frac{x^3 dx}{x^4(x^4 - 1)}$.Let $u = x^4$,then $du

Vedclass · Accepted Answer

$\frac{1}{4}\ln \left| \frac{x^4 - 1}{x^4} \right| + C$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{x(x^4 - 1)}$.Multiply the numerator and denominator by $x^3$:$I = \int \frac{x^3 dx}{x^4(x^4 - 1)}$.Let $u = x^4$,then $du = 4x^3 dx$,which implies $x^3 dx = \frac{du}{4}$.Substituting these into the integral:$I = \frac{1}{4} \int \frac{du}{u(u - 1)}$.Using partial fractions:$\frac{1}{u(u - 1)} = \frac{1}{u - 1} - \frac{1}{u}$.Therefore,$I = \frac{1}{4} \int \left( \frac{1}{u - 1} - \frac{1}{u} \right) du$.$I = \frac{1}{4} (\ln |u - 1| - \ln |u|) + C$.$I = \frac{1}{4} \ln \left| \frac{u - 1}{u} \right| + C$.Substituting $u = x^4$ back:$I = \frac{1}{4} \ln \left| \frac{x^4 - 1}{x^4} \right| + C$.

$\int \frac{dx}{x(x^4 - 1)}$ is equal to :-

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