= Solution
For a <continuously differentiable> <diffeomorphism> $\phi:J\to I$ between <intervals>, the <integration by substitution> states
$$
\int_I h(x)\,dx=\int_J h(\phi(t))|\phi'(t)|\,dt.
$$
It holds as an <identity element> of extended <integrals> for nonnegative <measurable> $h$. For signed or complex-valued <measurable> $h$, <integrability> on one side is equivalent to <integrability> after <substitution>, and the <identity element> then holds. The absolute <derivative> accounts for a decreasing <substitution>; on oriented finite <intervals> the equivalent formula uses $\phi'$ and oriented endpoints.
Put $C=\int_0^\infty(1+x^4)^{-1}\,dx=\pi/(2\sqrt2)$. The map $x=y^{-1/4}$ is a decreasing <diffeomorphism> of the positive half-line onto itself and
$$
\frac{|dx/dy|}{1+x^4}=\frac1{4y^{1/4}(1+y)}.
$$
Applying the nonnegative version before assuming finiteness proves both existence and value:
$$
\boxed{\int_0^\infty\frac{dy}{y^{1/4}(1+y)}=4C=\sqrt2\pi.}
$$
<Tonelli's theorem> permits either iterated <integral> of a nonnegative <measurable function> on a product of sigma-finite <measure spaces> and identifies both with its extended product <integral>. <Fubini's theorem> states that for an absolutely <integrable> <measurable function> on such a product, almost every section is <integrable>, the section-integral functions are <integrable>, and the two iterated <integrals> equal the product <integral>.
The given two-variable function is <continuous> and positive on the positive quadrant. For fixed $y>0$, substituting $t=xy^{1/4}$ gives
$$
\int_0^\infty f(x,y)\,dx=\frac{Cy^{-1/4}}{1+y}.
$$
Tonelli now gives
$$
\iint f(x,y)\,dx\,dy=C(4C)=4C^2=\frac{\pi^2}2<\infty.
$$
This proves <absolute integrability>, so Fubini is applicable.
For $x\neq1$, integrate the displayed partial fraction <identity element> in $y$ up to $R$:
$$
\int_0^R f(x,y)\,dy=\frac{\log(1+x^4R)-\log(1+R)}{x^4-1}
\longrightarrow\frac{4\log x}{x^4-1}.
$$
At $x=1$ the section <integral> is $\int_0^\infty(1+y)^{-2}\,dy=1$, agreeing with the <continuous> limit of this expression. Therefore Fubini identifies the remaining <integral> and gives
$$
\boxed{\int_0^\infty\frac{\log x}{x^4-1}\,dx=\frac14\cdot\frac{\pi^2}2=\frac{\pi^2}8.}
$$
The apparent <singularity> at one is removable, with limiting integrand value $1/4$. Positivity and the finite product <integral> already establish convergence at both ends.
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