For x 1,evaluate the integral: int frac{1}{ | vedclass

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(B) To solve the integral $I = \int \frac{1}{x(x^4 - 1)} \, dx$,we multiply the numerator and denominator by $x^3$: $I = \int \frac{x^3}{x^4(x^4 - 1)}

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$\frac{1}{4} \log \left( \frac{x^4 - 1}{x^4} \right) + K$

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(B) To solve the integral $I = \int \frac{1}{x(x^4 - 1)} \, dx$,we multiply the numerator and denominator by $x^3$: $I = \int \frac{x^3}{x^4(x^4 - 1)} \, dx$. Let $u = x^4$,then $du = 4x^3 \, dx$,which implies $x^3 \, dx = \frac{du}{4}$. Substituting these into the integral: $I = \frac{1}{4} \int \frac{1}{u(u - 1)} \, du$. Using partial fractions: $\frac{1}{u(u - 1)} = \frac{1}{u - 1} - \frac{1}{u}$. $I = \frac{1}{4} \int \left( \frac{1}{u - 1} - \frac{1}{u} \right) \, du = \frac{1}{4} [\log |u - 1| - \log |u|] + K$. $I = \frac{1}{4} \log \left| \frac{u - 1}{u} \right| + K$. Substituting $u = x^4$ back: $I = \frac{1}{4} \log \left( \frac{x^4 - 1}{x^4} \right) + K$.

For $x > 1$,evaluate the integral: $\int \frac{1}{x(x^4 - 1)} \, dx$

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