Alternating series involving zeta function

Take the logarithm of the Weierstass product form of the Gamma function $$ \frac{1}{\Gamma(x)}=xe^{\gamma x}\prod_{n=1}^\infty \left( 1+\frac{x}{n}\right)e^{-x/n}$$ to obtain $$-\log\Gamma(x)=\log x+\gamma x+\sum_{n=1}^\infty \log\left(1+\frac{x}{n}\right)-\frac{x}{n}$$ Now, $$\sum_{n=1}^\infty \log\left(1+\frac{x}{n}\right)-\frac{x}{n} =-\sum_{n=1}^\infty\sum_{m=2}^\infty\frac{\left(-\frac{x}{n}\right)^m}{m} \\\\=-\sum_{m=2}^\infty\frac{(-1)^m x^m}{m}\sum_{n=1}^\infty\frac{1}{n^m}=-\sum_{m=2}^\infty\frac{(-1)^m x^m \zeta(m)}{m}$$ so that $$\sum_{n=2}^\infty\frac{(-1)^n \zeta(n)}{n}x^n=\gamma x+\log(x)+\log\Gamma(x)$$ and $$\sum_{n=2}^\infty\frac{(-1)^n \zeta(n)}{n(n+1)}=\int_{0}^{1}(\gamma x+\log(x)+\log\Gamma(x))dx=\frac{\gamma}{2}-1+\int_{0}^{1}\log\Gamma(x)dx$$ $\int_{0}^{1}\log\Gamma(x)dx=\log\sqrt{2\pi}$ is quite famous. check here for reduction to evaluating $\int_0^\pi \log\sin(x)dx$ and here for the evaluation