Prove $\int_0^{\infty}\frac{\ln (x)}{(x^2+1)(x^3+1)}\ dx=-\frac{37}{432}\pi^2$ with real method

A real-analytic technique may be to exploit $$ I = \int_0^1 \frac{1-x^3}{(1+x^2)(1+x^3)}\log(x)\,dx $$ then expand $f(x)=\frac{1-x^3}{(1+x^2)(1+x^3)}$ as a Taylor series around $x=0$, $$ f(x) = \sum_{n\geq 0}\left(1-x^2-2 x^3+x^4+2 x^5+x^6-2 x^7-x^8 + x^{10} \right) x^{12n} $$ and exploit $$ \int_0^1 x^k\log(x)\,dx = -\frac{1}{(k+1)^2} $$ to convert $I$ into a combination of Dirichlet $L$-functions $L(s,\chi)$ where $s=2$ and $\chi$ is a Dirichlet character $\!\!\pmod{12}$, or a combination of trigamma functions evaluated at multiples of $\frac{1}{12}$: the reflection formula for $\psi'$ is helpful, since it gives $$ \sum_{n\geq 0}\left(\frac{1}{(12n+1)^2}+\frac{1}{(12n+11)^2}\right) = \frac{2+\sqrt{3}}{36}\,\pi^2,$$ $$ \sum_{n\geq 0}\left(\frac{1}{(12n+3)^2}+\frac{1}{(12n+9)^2}\right)=\frac{1}{72}\,\pi^2,$$ $$ \sum_{n\geq 0}\left(\frac{1}{(12n+4)^2}+\frac{1}{(12n+8)^2}\right)=\frac{1}{108}\,\pi^2,$$ $$ \sum_{n\geq 0}\left(\frac{1}{(12n+5)^2}+\frac{1}{(12n+7)^2}\right)=\frac{2-\sqrt{3}}{36}\,\pi^2,$$ $$ \sum_{n\geq 0}\frac{1}{(12n+6)^2}=\frac{1}{288}\pi^2.$$ To deduce $I=\color{red}{-\frac{37}{432}\pi^2}$ is now just a matter of simple algebra.

Preliminary Result:

$\displaystyle \frac{2}{(x^2+1)(x^3+1)} \equiv \frac{x+1}{x^2+1}-\frac{x^2+x-1}{x^3+1}$

Proof: Obvious.

Consider the parametrised integral:

$\displaystyle f(\alpha) = \int \frac{x^\alpha \, \text{d}x}{(x^2+1)(x^3+1)} = \frac{1}{2} \int \frac{x^{\alpha+1}+x^\alpha}{x^2+1}-\frac{x^{\alpha+2}+x^{\alpha+1}-x^\alpha}{x^3+1} \, \text{d}x$

For the first part, substitute $t = x^2$, and $t = x^3$ for the second part

$\displaystyle f(\alpha) = \frac{1}{4} \int_0^\infty \frac{t^{\frac{\alpha}{2}}+t^\frac{\alpha-1}{2}}{1+t} \text{d}t +\frac{1}{6}\int_0^\infty \frac{t^{\frac{\alpha}{3}}+t^{\frac{\alpha-1}{3}}-t^{\frac{\alpha-2}{3}}}{1+t} \text{d}t$

Use the Integral Representations of the Beta Function to obtain:

$\displaystyle f(\alpha) = \frac{\text{B}(1+\frac{\alpha}{2},-\frac{\alpha}{2})+\text{B}(\frac{1+\alpha}{2},\frac{1-\alpha}{2})}{4} - \ \frac{\text{B}(1+\frac{\alpha}{3},-\frac{\alpha}{3})+\text{B}(\frac{2+\alpha}{3},\frac{1-\alpha}{3})-\text{B}(\frac{1+\alpha}{3},\frac{2-\alpha}{3})}{6}$

$\displaystyle \frac{f(\alpha)}{\pi} = \frac{\csc{\left(\pi+\frac{\pi \alpha}{2} \right)}+\csc{\left(\frac{\pi}{2}+\frac{\pi \alpha}{2}\right)}}{4}-\frac{\csc{\left( \pi + \frac{\pi \alpha}{3} \right)}+\csc{\left(\frac{2\pi}{3} +\frac{\pi \alpha}{3}\right)}-\csc{\left( \frac{\pi}{3} + \frac{\pi \alpha}{3}\right)}}{6}$

$\displaystyle \frac{12f(\alpha)}{\pi} = 3\sec{\frac{\pi \alpha}{2}} -3\csc{\frac{\pi \alpha}{2}}+2\csc{\frac{\pi \alpha}{3}}+2\csc{\left( \frac{2\pi}{3}+\frac{\pi \alpha}{3}\right)}-2\csc{\left(\frac{\pi}{3}+\frac{\pi \alpha}{3}\right)}$

Differentiate both sides with respect to $\alpha$ and evaluate at $0$ to obtain the integral you want.

Let's define the branch of $\log$ to lie on the posive real axis. We consider the complex valued function

$$ f(z)=\frac{\log(z)^2}{(z^2+1)(z^3+1)} $$

we integrate this function around a keyhole contour in the complex plane with slit at the positive real axis. The contributions from the big circle will vanish out because $f(|z|)\sim \log(|z|)/|z|^5$ as $|z|\rightarrow \infty$ so we are left with

$$ \oint f(z)=\int_0^{\infty}\frac{\log(x)^2}{(x^2+1)(x^3+1)}-\int_0^{\infty}\frac{(\log(x)+2 i \pi)^2}{(x^2+1)(x^3+1)}=2\pi i\sum_i\text{Res}(f(z),z=z_i) $$

where are the roots of the denominator of $f(z)$: $z_0=-1,z_{2/3}=\pm i,z_{4,5}=e^{\pm i2 \pi/3}$. We can rewrite this as $$ \int_0^{\infty}\frac{\log(x)}{(x^2+1)(x^3+1)}=\sum_i\text{Res}(f(z),z=z_i)-2\pi i\int_0^{\infty}\frac{1}{(x^2+1)(x^3+1)} $$

The remaining integral can be evaluated either by cumbersome partial fraction decomposition or by considering

$$ g(z)=\frac{\log(z)}{(z^2+1)(z^3+1)} $$

integrated around the same contour then above.

I leave the details to you but from now on (also the calculation of residues if one keep in mind $\log(i)=\frac{i\pi}{2},\log(i)=\frac{3i\pi}{2}$ for this choice of branchcut) so i leave the details to you...

As an alternative the remaining integral might be ignored because it will yield a purely imaginary contribution and we can write (the lhs is purely real)

$$ \int_0^{\infty}\frac{\log(x)}{(x^2+1)(x^3+1)}=\Re\left[\sum_i\text{Res}(f(z),z=z_i)\right] $$

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This method will work for

$$ \int_0^{\infty}\frac{\log(z)^2}{(z^2+1)(z^n+1)}dz $$

as long $n$ is a natural number