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We make the change of variables $x = -w$, $dx = -dw$, and change the limits of integration to obtain

$\begin{eqnarray*} \int_{-\infty}^0 \frac{x^2}{1+x^4} \log|x| \; dx &=& -\int_0^\infty \frac{w^2}{1+w^4} \log|-w| \; dw \\ &=& -\int_0^\infty \frac{w^2}{1+w^4} \log w \; dw \end{eqnarray*}$

But this is clearly wrong, since $\frac{x^2}{1+x^4} \log|x|$ is an even function. Where is the mistake?

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    After the substitution you get the integral $\int_{\infty}^0\dots$. Flipping the limits gives you another negative sign.2012-05-17
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    I see. Thanks you!2012-05-17
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    @MihaHabič Please do post this as a solution.2012-05-17

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