Interesting Zeta

First of all note the series $\displaystyle \psi(z+1)+\gamma=\sum_{k=1}^{\infty} (-1)^{k+1}\zeta(k+1) z^{k}$ For $\vert z\vert\lt 1$ Now multiplying both sides by $z$ and then substituting $z=-x$ we get $\displaystyle -x\psi(1-x)-\gamma x=\sum_{k\ge 2} \zeta(k)x^k$ Now we see that $\displaystyle \xi=\sum_{k\ge 2} \frac {\zeta(k)-1}{k+1}=\sum_{k\ge 2} (\zeta(k)-1)\int_0^1 x^k dx$ Interchanging summation and integration which can be justified by Tonelli's Theorem we get $\displaystyle \int_0^1 \left(\left(\sum_{k\ge 2} \zeta(k)x^k\right)-\sum_{k\ge 2} x^k\right)dx$

Using the above obtained result alongwith the formula for infinite geometric progression we get $\xi=\int_0^1 \left(-x\psi(1-x)-\gamma x-\frac {x^2}{1-x}\right)dx$ $\xi=\int_0^1 \left(-x\psi(1-x)-\gamma x-\left(-x-1+\frac {1}{1-x}\right)\right)dx$

Now properly splitting the integrals we have $\xi=\xi_1+\xi_2$

Where $\displaystyle \xi_1=-\int_0^1 \left(x\psi(1-x)+\frac {1}{1-x}\right)$ And $\displaystyle \xi_2=\int_0^1 (x+1-\gamma x)dx=\color{#D61F06}{\frac 32-\frac {\gamma}{2}}$

In $\xi_1$ we use the property that $\displaystyle \int_a^b f(x)dx=\int_a^b f(a+b-x)dx$ to get $\displaystyle \xi_1=-\int_0^1\left((1-x)\psi(x)+\frac 1x\right)dx=-\int_0^1 \left(\psi(x)+\frac 1x\right)dx +\int_0^1 x\psi(x)dx$

Now using the functional rule for digamma function that $\psi(z+1)=\psi(z)+\frac 1z$ we get $\displaystyle \xi_1=-\int_0^1 \psi(x+1) dx +\int_0^1 x\psi(x)dx=\left[\ln(\Gamma(x+1))\right]_0^1+\int_0^1 x\psi(x) dx=\int_0^1 x\psi(x) dx$

So now $\xi_1=\int_0^1 x\psi(x) dx$ Now using IBP once we get $\displaystyle \xi_1=-\int_0^1 \ln(\Gamma(x)) dx=\color{#D61F06}{-\frac 12\ln(2\pi)}$ Where I use the standard result $\displaystyle \int_0^1 \ln(\Gamma(x)) dx=\frac 12\ln(2\pi)$ Which has been asked on Brilliant earlier also(I couldn't find the link to it).

So at the end we get $\displaystyle \xi=\xi_1+\xi_2=\frac 32-\frac 12\ln(2\pi)-\frac {\gamma}{2}$

Thus $A+B+C+D=\boxed {8}$