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$$\int\dfrac{x^2}{1+x^4}dx$$ I tried many standard approaches, but I didn't get too far! Here's the most promising of them: $$\int\dfrac{dx}{\frac{1}{x^2}+x^2}$$ knowing that $\left(1/x+x\right)^2=\frac{1}{x^2}+x^2+2$ we can change variables $1/x+x=t$. Unluckily this doesn't work either.

José Carlos Santos
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asd11
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    $x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2=(x^2+1+\sqrt{2}x)(x^2+1-\sqrt{2}x)$. Therefore you can write $\frac{x^2}{x^4+1}=\frac{Ax+B}{x^2+\sqrt{2}x+1}+\frac{Cx+D}{x^2-\sqrt{2}x+1}$ – Hellen Jul 22 '17 at 11:54
  • Thank you. That is revealing! – asd11 Jul 22 '17 at 11:56

3 Answers3

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HINT:

Like $\int \frac{x^2}{x^4+x^2+1}\ dx$,

$$\dfrac{2x^2}{x^4+1}=\dfrac{1-1/x^2}{x^2+1/x^2}+\dfrac{1+1/x^2}{x^2+1/x^2}$$

lab bhattacharjee
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Hint: Express $$\frac{x^2}{1+x^4}=\frac{ax+b}{x^2-\sqrt{2}x+1}+\frac{cx+d}{x^2+\sqrt{2}x+1}$$ and find $a,b,c,d$.

ntt
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Hint:$$\frac { { x }^{ 2 } }{ 1+{ x }^{ 4 } } =\frac { { x }^{ 2 } }{ \left( 1-{ ix }^{ 2 } \right) \left( 1+{ ix }^{ 2 } \right) } =\frac { 1 }{ 2i } \left[ \frac { 1 }{ 1-{ ix }^{ 2 } } -\frac { 1 }{ 1+i{ x }^{ 2 } } \right] \\ $$

haqnatural
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