int frac{x^2}{(x^2-1)(x^2+1)} dx = | vedclass

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(B) Let $I = \int \frac{x^2}{(x^2-1)(x^2+1)} dx$. Using partial fractions,we can write $\frac{x^2}{(x^2-1)(x^2+1)} = \frac{1}{2} \left( \frac{1}{x^2-1

Vedclass · Accepted Answer

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

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(B) Let $I = \int \frac{x^2}{(x^2-1)(x^2+1)} dx$. Using partial fractions,we can write $\frac{x^2}{(x^2-1)(x^2+1)} = \frac{1}{2} \left( \frac{1}{x^2-1} + \frac{1}{x^2+1} ight)$. Now,$I = \frac{1}{2} \int \frac{1}{x^2-1} dx + \frac{1}{2} \int \frac{1}{x^2+1} dx$. Using the standard integrals $\int \frac{1}{x^2-a^2} dx = \frac{1}{2a} \ln \left| \frac{x-a}{x+a} ight| + c$ and $\int \frac{1}{x^2+1} dx = 	an^{-1} x + c$,we get: $I = \frac{1}{2} \left( \frac{1}{2} \ln \left| \frac{x-1}{x+1} ight| ight) + \frac{1}{2} 	an^{-1} x + c$. $I = \frac{1}{4} \ln \left| \frac{x-1}{x+1} ight| + \frac{1}{2} 	an^{-1} x + c$.

$\int \frac{x^2}{(x^2-1)(x^2+1)} dx = $

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