<s>[INST] Prove $\int_0^{\infty}\frac{\ln (x)}{(x^2+1)(x^3+1)}\ dx=-\frac{37}{432}\pi^2$ with real method I came across the following integral:
$$\large{\int_0^\infty \frac{\ln (x)}{(x^2+1)(x^3+1)}\ dx=-\frac{37}{432}\pi^2}$$
I know it could be solved with resuide method, and I want to know if there are some real methods can sove it?
Meanwhile,I remember a similar integral:
$$\large{\int_0^\infty \frac{1}{(x^2+1)(x^a+1)}\ dx=\frac{\pi}{4}}$$
And I want to know the following one:
$${\color{red}{\large{\int_0^\infty \frac{\ln x}{(x^2+1)(x^a+1)}\ dx = \huge{?}}}}$$
Using the Mathematica I got the follow result.

Could you suggest some ideas how to prove this?  Any hints will be appreciated.
 [/INST] 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.
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