How to show that: $$ \int_{0}^{\infty}\frac{1}{e^x-1}\left[\frac{12\,e^x}{(e^x-1)^2}-\frac{12}{x^2}+1\right]\,dx=\frac{5}{2}+\frac{\zeta'(2)}{\zeta(2)}-\frac{\zeta'(0)}{\zeta(0)}\tag{1} $$ Is it possible to split the integral into two integrals representing ${\zeta'}/{\zeta}(0)$ and ${\zeta'}/{\zeta}(2)$.
The logarithmic derivative of Riemann Zeta function is mainly defined by the infinte series:
$$ \log'\left[\zeta(s)\right]=\frac{\zeta'}{\zeta}(s)=-\sum_{n=1}^{\infty}\frac{\Lambda(n)}{n^s} $$
Where $\Lambda(n)$ is Von Mangoldt function. ${\zeta'}/{\zeta}$ has some intresting idendities, like:
$$ \frac{\zeta'}{\zeta}(s)+\frac{\zeta'}{\zeta}(1-s)=\log\pi-\frac{1}{2}\frac{\Gamma'}{\Gamma}\left(\frac{s}{2}\right)-\frac{1}{2}\frac{\Gamma'}{\Gamma}\left(\frac{1-s}{2}\right) $$
Does ${\zeta'}/{\zeta}$ has an integral representation? Thanks.