int frac{x}{x^4 - 1} dx = | vedclass

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Question

(A) Let $I = \int \frac{x}{x^4 - 1} dx$. Substitute $t = x^2$,then $dt = 2x dx$,which implies $x dx = \frac{1}{2} dt$. Substituting these into the int

Vedclass · Accepted Answer

$\frac{1}{4} \log \left| \frac{x^2 - 1}{x^2 + 1} \right| + c$

Vedclass · Answer

(A) Let $I = \int \frac{x}{x^4 - 1} dx$. Substitute $t = x^2$,then $dt = 2x dx$,which implies $x dx = \frac{1}{2} dt$. Substituting these into the integral: $I = \frac{1}{2} \int \frac{1}{t^2 - 1} dt$. Using the standard formula $\int \frac{1}{t^2 - a^2} dt = \frac{1}{2a} \log \left| \frac{t - a}{t + a} \right| + c$: $I = \frac{1}{2} \left( \frac{1}{2(1)} \log \left| \frac{t - 1}{t + 1} \right| \right) + c$. $I = \frac{1}{4} \log \left| \frac{x^2 - 1}{x^2 + 1} \right| + c$.

$\int \frac{x}{x^4 - 1} dx = $

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