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The question asks to evaluate the integral: $$\int_0^\infty \frac{\ln^2{x}}{1+x^4} \mathrm{d}x.$$

I have tried a few substitutions but am not getting anywhere.

Thanks in advance!

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Assume $0<s<4$. We start with the following general result

$$ \int_0^\infty \frac{x^{s-1}}{1+x^4}\mathop{dx}=\frac{\pi}4\:\frac1{\sin\left(\pi s/4\right)}. \tag1 $$

Proof. By performing the change of variable $u=\dfrac{1}{1+x^4}$, then using the Euler beta function $B$, we have $$ \begin{align} \int_0^\infty \frac{x^{s-1}}{1+x^4}\mathop{dx}&=\frac14\int_0^1 u^{\large 1-\frac{s}4-1}(1-u)^{\large \frac{s}4-1}\mathop{du}\\\\ &=\frac14B\left(1-\frac s4,\frac s4\right)\\\\ &=\frac14\Gamma\left( 1-\frac{s}4\right)\Gamma\left(\frac{s}4\right)\\\\ &=\frac{\pi}4\:\frac1{\sin\left(\pi s/4\right)} \end{align} $$ giving $(1)$, where we have used the reflection formula for the $\Gamma$ function.

Then one may differentiate $(1)$ twice getting $$ \int_0^\infty \frac{x^{s-1}(\ln x)^2}{1+x^4}\mathop{dx}=\frac{\pi}4\left(\frac1{\sin\left(\pi s/4\right)}\right)_{\large s}'' \tag2 $$ then put $s:=1$ to obtain

$$ \int_0^\infty \frac{\ln^2 x}{1+x^4}\mathop{dx}=\frac{3\pi^3}{64}\sqrt{2}. \tag3 $$

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Note \begin{eqnarray} \int_0^\infty \frac{\ln^2{x}}{1+x^4} \mathrm{d}x&=&\int_0^1 \frac{\ln^2{x}}{1+x^4} \mathrm{d}x+\int_1^\infty \frac{\ln^2{x}}{1+x^4} \mathrm{d}x\\ &=&\int_0^1 \frac{(1+x^2)\ln^2{x}}{1+x^4} \mathrm{d}x\\ &=&\frac{3\pi^3}{64}\sqrt2. \end{eqnarray} See the result in this post.

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