(Okay, this integral made me go on a complex analysis goose chase that eventually trolled me smh with an integral that integrated through an asymptote.)
I've come to the conclusion that complex analysis won't really work here, so let's use series.
We seek to find $$I=\int_1^{\infty}\frac{\ln(x)}{x^2+1}\text{ d}x$$
Performing a change of variables, we have
$$u=\ln(x),\qquad\frac{\text{d}u}{\text{d}x}=\frac1x,\qquad\text{d}x=x\text{ d}u,\qquad x=e^u$$
$$\implies\int_0^\infty\frac{u\cdot e^u}{e^{2u}+1}\text{ d}u=\int_0^\infty\frac{u}{e^{u}+e^{-u}}\text{ d}u=\int_0^\infty\frac{x\cdot e^{-x}}{1+e^{-2x}}\text{ d}x$$
I just rewrote the $u$ into $x$ at the end just for convenience.
Now let's take a look at the general case where we have
$$\int_0^\infty \frac{x^{s-1}\cdot e^{-x}}{1+e^{-2x}}\text{ d}x$$
We can see that
\begin{align}
\int_0^\infty x^{s-1}\cdot\left(\frac{e^{-x}}{1-(-e^{-2x})}\right)\text{ d}x&=\int_0^\infty x^{s-1}\cdot\left(\sum_{n=0}^\infty (-1)^ne^{-(2n+1)x}\right)\text{d}x\\
&=\sum_{n=0}^\infty (-1)^n\int_0^\infty x^{s-1} e^{-(2n+1)x}\text{ d}x\\
&=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}\cdot\Gamma(s)\\
&=\beta(s)\Gamma(s)
\end{align}
where the first line follows from a geometric series expansion, the second from the definition of the gamma function, and the third from the definition of the Dirichlet beta function.
In our case, we have $s=2$ So $$I=\beta(2)\Gamma(2)$$
In particular, we see that $\Gamma(2)=1!=1$ but more importantly, $\beta(2)=C$ because by definition the Catalan constant $C$ is equal to the series representation of the Dirichlet beta function with parameter 2.
Hence we have shown that $$I=\boxed{C}$$
