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$\int \frac{x^2}{\left(x^2-1\right)\left(x^2+1\right)} d x=$
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Verified Answer
The correct answer is:
$\frac{1}{4} \log \left|\frac{x-1}{x+1}\right|+\frac{1}{2} \operatorname{Tan}^{-1} x+c$
Let $\int \frac{x^2}{\left(x^2-1\right)\left(x^2+1\right)} d x=I$
$\Rightarrow I=\frac{1}{2} \int\left(\frac{1}{x^2-1}+\frac{1}{x^2+1}\right) d x$
$\begin{aligned} & =\frac{1}{2} \cdot \frac{1}{2} \ln \left|\frac{x-1}{x+1}\right|+\frac{1}{2} \tan ^{-1} x+c \\ & =\frac{1}{4} \ln \left|\frac{x-1}{x+1}\right|+\frac{1}{2} \tan ^{-1} x+c\end{aligned}$
$\Rightarrow I=\frac{1}{2} \int\left(\frac{1}{x^2-1}+\frac{1}{x^2+1}\right) d x$
$\begin{aligned} & =\frac{1}{2} \cdot \frac{1}{2} \ln \left|\frac{x-1}{x+1}\right|+\frac{1}{2} \tan ^{-1} x+c \\ & =\frac{1}{4} \ln \left|\frac{x-1}{x+1}\right|+\frac{1}{2} \tan ^{-1} x+c\end{aligned}$
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