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.

Here is a variant of @tired's solution. In this solution, you are only required to know:


*

*How to expand a rational function into partial fraction decomposition,

*How to compute complex logarithm, and

*The following claim:



Claim. For any $z \in \Bbb{C}\setminus(-\infty, 0]$ we have
  $$ I(z) := \int_{0}^{\infty} \left( \frac{\log x}{x+z} - \frac{\log x}{x+1} \right) \, dx = -\frac{1}{2}\log^2 z \tag{1}$$
  where $\log$ is the principal value of the complex logarithm.


Computation. Before proving the claim, let us see how $\text{(1)}$ allows us to compute the integral in question. Using the partial fraction decomposition, we get
\begin{align*}
\frac{1}{(1+x^2)(1+x^3)}
&= \frac{1-i}{4} \cdot\frac{1}{x-i} + \frac{1+i}{4}\cdot\frac{1}{x+i} \\
&\qquad \qquad + \frac{1}{6}\cdot \frac{1}{x+1} - \frac{1}{3}\cdot \frac{1}{x - e^{\pi i/3}} - \frac{1}{3}\cdot \frac{1}{x - e^{-\pi i/3}}.
\end{align*}
(Obviously this is a disguise of @tired's complex-analytic solution, since this decomposition often comes from residue computation. On the other hand, residue is no longer a necessity and thus one can work purely with algebra.)
Also notice that the sum of 'coefficients' are zero: $ \frac{1-i}{4} + \frac{1+i}{4} + \frac{1}{6} - \frac{1}{3} - \frac{1}{3} = 0$. Therefore it follows from the claim that
\begin{align*}
\int_{0}^{\infty} \frac{\log x}{(1+x^2)(1+x^3)} \, dx
&= \frac{1-i}{4} I(-i) + \frac{1+i}{4}I(i)\\
&\qquad \qquad + \frac{1}{6}I(1) - \frac{1}{3}I(-e^{\pi i/3}) - \frac{1}{3}I(-e^{-\pi i/3}) \\
&= -\frac{1}{2} \bigg[ \frac{1-i}{4} \left( -\frac{i\pi}{2} \right)^2 + \frac{1+i}{4}I\left( \frac{i\pi}{2} \right)^2 \\
&\qquad \qquad + 0 - \frac{1}{3}\left( -\frac{2i\pi}{3} \right)^2 - \frac{1}{3}\left( \frac{2i\pi}{3} \right)^2 \bigg] \\
&= -\frac{37}{432}\pi^2.
\end{align*}

Proof of Claim. Now it remains to compute the integral $\text{(1)}$ This is easily done by differentiating $I(z)$:
$$ I'(z) = - \int_{0}^{\infty} \frac{\log x}{(x+z)^2} \, dx. $$
In order to compute this integral, we first replace the lower limit by $\epsilon > 0$ to obtain
$$ - \int_{\epsilon}^{\infty} \frac{\log x}{(x+z)^2} \, dx = \frac{\epsilon \log \epsilon}{z(z+\epsilon)} - \frac{1}{z}\log (z+\epsilon). $$
Taking limit as $\epsilon \to 0^+$, we get
$$ I'(z) = -\frac{1}{z}\log z, \qquad I(1) = 0. $$
This is enough to prove the claim. ////
