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Integrate $$\int\frac{x^2dx}{1+x^4}$$ I've factored the denominator to $(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$ and got stuck.

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  • $\begingroup$ it is good starting point,i will upload picture $\endgroup$ – dato datuashvili Apr 2 '13 at 19:59
  • $\begingroup$ ok i have deleted my answer.first of all when i am posting answer from wolfram alpha,because it requires some autoriation like facebook or registration,that why maybe OP is not registered on facebook,he should worry about this.but no problem .i have deleted my asnwer $\endgroup$ – dato datuashvili Apr 2 '13 at 20:06
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Or, use $$\frac{x^2}{1+x^4}=\frac{1}{2} \frac{2x^2}{1+x^4}=\frac{1}{2}\left(\frac{x^2-1}{1+x^4}+\frac{x^2+1}{1+x^4}\right)$$ and this idea.

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The next step is to perform a partial fraction decomposition: $$ \frac{x^2}{(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)} = \frac{1}{2 \sqrt{2}} \frac{x}{x^2-\sqrt{2}x+1} - \frac{1}{2 \sqrt{2}} \frac{x}{x^2+\sqrt{2}x+1} $$ And the than use the table anti-derivative for $\int \frac{x}{x^2 + a x+b} \mathrm{d}x$.

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