I encountered following two integrals $$ \begin{aligned} &I_1=\int_{0}^{\infty} \frac{\ln x}{(x+1)(x+2)(\pi^2+\ln^2x)} \text{d}x\\ &I_2=\int_{0}^{\infty} \frac{\ln (1+x)}{(x+1)(x+2)(\pi^2+\ln^2x)} \text{d}x \end{aligned} $$ The first one was easier.By contour integration we might have $$ I_1 =\gamma+\text{Re}\left [ \psi^{(0)}\left (\frac{\ln 2}{2\pi i} \right ) \right ] $$ But $I_2$ is individual.Take the substitution $x\to e^{x}$.Then $I_2$ is equal to $$ I_2=\int_{-\infty}^{\infty} \frac{e^x\ln(1+e^x)}{(e^x+1)(e^x+2)} \frac{\text{d}x}{\pi^2+x^2} $$ Now I am stuck on this.(My first attempt is too long.) Any suggestion is appreciated.
Update: Using $$ \int_{0}^{1}\frac{1}{t+x} \text{d}t =\ln\left ( 1+\frac{1}{x} \right ) $$ Thus we have $$ I_2 = \int_{0}^{1} \left [ \frac{1}{(t-1)(2t-1)} \left ( \frac{1}{\ln t} -\frac{1}{t-1} \right ) +4\left ( 1-\frac{1}{2\ln2} \right ) \frac{1}{2t-1} -\frac{1}{2}\frac{1}{t-1} \right ] \text{d}t $$ or $$I_2= \int_{0}^{\infty}\left ( \frac{\frac{1}{x}-\frac{1}{1-e^{-x}} }{(e^x-1)(2e^{-x}-1)} +\frac{1}{2}\frac{1}{e^x-1} - \frac{4\left ( 1-\frac{1}{2\ln2} \right ) }{e^x-2} \right ) \text{d}x$$
Update2: First note that $$ I_2=\underbrace{\int_{0}^{\infty} \frac{\ln\left (1+ \frac{1}{x} \right ) } {(x+\frac{1}{2} )(\pi^2+\ln^2x)} \text{d}x} _{I_3} -\underbrace{\int_{0}^{\infty} \frac{\ln\left ( 1+\frac{1}{x} \right ) } {(x+1)(\pi^2+\ln^2x)} \text{d}x}_{I_4} $$ And $$ I_4=I_4=\frac{1}{2} \left(\ln2\pi-1-\gamma\right) $$ $$ I_3=-2\int_{0}^{\infty} \frac{\frac{1}{x}-\frac{1}{1-e^{-x}}+2-\frac{1}{\ln2} } {e^x-2} \text{d}x $$ And $$I_2=\frac{1}{2} \left(1+\gamma-\ln2\pi\right)-2\int_{0}^{\infty} \frac{\frac{1}{x}-\frac{1}{1-e^{-x}}+2-\frac{1}{\ln2} } {e^x-2} \text{d}x$$ Remark:This integral related to "Incomplete polygamma functions".It seems like a closed-form.
Corollary $$\int_{0}^{\pi/2} \frac{x\cot x\ln(\cos x)}{x^2+\ln^2(2\cos x)} \text{d}x -\pi\int_{0}^{\infty} \frac{\frac{1}{x}-\frac{1}{1-e^{-x}}+2-\frac{1}{\ln2} } {e^x-2} \text{d}x =\frac{\pi}{8} \left ( 3+\gamma-7\ln2+3\ln\pi +4\text{Re}\left [ \psi^{(0)}\left ( \frac{\ln2}{2\pi i} \right ) \right ] \right )$$