What to do with this Prove that
$$
\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x(x+1)}dx=\frac{1}{2} (\ln 2)^2-\ln \pi \ln2   
$$
I separated them 
$$
\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x}dx-\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x+1}dx
$$
For the former integral i tried to use differentiation under integration but got stuck and i have no idea about the latter one. Plz help!

\begin{align}
&\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x}dx-\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x+1}dx=\\
&\int^{1}_{0} \frac{\ln\frac{\sin \pi x}{2\sin(\pi x/2)}}{x}dx-\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x+1}dx=\\
&\lim_{\varepsilon\to 0}\left(\int^{1}_{\varepsilon} \frac{\ln\frac{\sin\pi x}{2}}{x}dx-\int^{1}_{\varepsilon} \frac{\ln \left(\sin \left(\frac{\pi x}{2} \right)\right)}{x}dx-\int^{1}_{0} \frac{\ln \left(\cos \left(\frac{\pi x}{2} \right)\right)}{x+1}dx\right)=\\
&\lim_{\varepsilon\to 0}\left(\int^{1}_{\varepsilon} \frac{\ln\frac{\sin\pi x}{2}}{x}dx-\int^{1}_{\varepsilon} \frac{\ln \left(\sin \left(\frac{\pi x}{2} \right)\right)}{x}dx-\int^{2}_{1} \frac{\ln \left(\sin \left(\frac{\pi x}{2} \right)\right)}{x}dx\right)=\\
&\lim_{\varepsilon\to 0}\left(\int^{1}_{\varepsilon} \frac{\ln\frac{\sin\pi x}{2}}{x}dx-\int^{2}_{\varepsilon} \frac{\ln \left(\sin \left(\frac{\pi x}{2} \right)\right)}{x}dx\right)=\\
&\lim_{\varepsilon\to 0}\left(\int^{1}_{\varepsilon} \frac{\ln\frac{\sin\pi x}{2}}{x}dx-\int^{1}_{\varepsilon/2} \frac{\ln \left(\sin \pi x\right)}{x}dx\right)=\\
&\lim_{\varepsilon\to 0}\left(-\ln 2\int^{1}_{\varepsilon}\frac{dx}{x}-\int^{\varepsilon}_{\varepsilon/2} \frac{\ln \left(\sin \pi x\right)}{x}dx \right)=\\
&\lim_{\varepsilon\to 0}\left(-\ln 2\int^{1}_{\varepsilon}\frac{dx}{x}-\int^{\varepsilon}_{\varepsilon/2} \frac{\ln \left(\pi x\right)}{x}dx \right)=\\
&\lim_{\varepsilon\to 0}\left(\ln 2\ln\varepsilon-\ln\pi\ln 2-\frac{1}{2}\left[(\ln\varepsilon)^2-(\ln\varepsilon/2)^2\right]\right)=\frac{1}{2} (\ln 2)^2-\ln \pi \ln2
\end{align}
