How to evaluate $\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{n+1}$ directly? The rational zeta series
$$
\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{n+1}=\frac{3}{2}-\ln \pi \tag1
$$
can be derived from other well known rational zeta series.
$$
\sum_{n=2}^{\infty}\frac{\left ( -1 \right )^{n}\left ( \zeta (n)-1 \right )}{n+1}=\frac{3}{2}+\frac{\gamma }{2}-\frac{\ln 8\pi}{2} \tag2
$$
$$
\sum_{n=2}^{\infty}\frac{\zeta (n)-1}{n+1}=\frac{3}{2}-\frac{\gamma }{2}-\frac{\ln 2\pi}{2} \tag3
$$
Zeta series (2) and (3) can be derived by integrating the Taylor series of logarithm of gamma function. Zeta series (2)+(3) gives
$$
\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{2n+1}=\frac{3}{2}-\frac{\ln 4\pi}{2} \tag4
$$
The zeta series below can be derived directly with the integral definition of $\zeta(2n)$.
$$
\sum_{n=1}^{\infty}\frac{\zeta (2n)}{(n+1)(2n+1)}=\frac{1}{2} \tag5
$$
From zeta series (5) we get
$$
\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{(2n+1)(2n+2)}=\frac{3}{4}-\ln 2 \tag6
$$
Zeta series (6) can be rewritten as
$$
\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{2n+1}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{n+1}=\frac{3}{4}-\ln 2 \tag7
$$
Tegether with zeta series (4), we get the result of zeta series (1).
Other than using known results of rational zeta series, how to evaluate zeta series (1) directly with elementary sum of series and integral?
I have tried several ways without success. One of my attempts:
$$
\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{n+1}x^{n+1}=\sum_{n=1}^{\infty}\sum_{k=2}^{\infty}\frac{1}{k^{2n}}\int_{0}^{x}t^{n}dt=\sum_{k=2}^{\infty}\int_{0}^{x}\sum_{n=1}^{\infty}\left ( \frac{t}{k^{2}} \right )^{n}dt \\
=\sum_{k=2}^{\infty}\int_{0}^{x}\frac{t}{k^{2}-t}dt=\sum_{k=2}^{\infty}\left ( k^{2}\ln\frac{k^{2}}{k^{2}-x}-x\right )
$$
It seems this attempt won't give any useful result for a closed form, although the sum of this series does converge to $(3/2-\ln\pi)$ slowly when setting $x=1$.

We write
$$ S := \sum_{n=1}^{\infty} \frac{\zeta(2n)-1}{n+1} $$
for the sum to be computed.

1st Solution. We have
\begin{align*}
S
= \sum_{n=1}^{\infty} \frac{1}{n+1} \sum_{k=2}^{\infty} \frac{1}{k^{2n}} 
= \sum_{k=2}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n+1} \frac{1}{k^{2n}} 
= \sum_{k=2}^{\infty} \left( - k^2 \log \left( 1 - \frac{1}{k^2} \right) - 1 \right).
\end{align*}
In order to compute this, we write $S_K$ for the partial sums of the last step. Then
\begin{align*}
S_K
&= -K + 1 + \sum_{k=2}^{K} k^2 \log \left( \frac{k^2}{(k+1)(k-1)} \right) \\
&= -K + 1 + \sum_{k=2}^{K} 2 k^2 \log k - \sum_{k=3}^{K+1} (k-1)^2 \log k - \sum_{k=1}^{K-1} (k+1)^2 \log k \\
&= -K + 1 + \log 2 - K^2 \log(K+1) + (K+1)^2 \log K \\
&\quad + \sum_{k=2}^{K} (2 k^2 - (k-1)^2 - (k+1)^2 ) \log k \\
&= -K + 1 + \log 2 - K^2 \log\left(1 + K^{-1}\right) + (2K+1)\log K - 2 \log (K!).
\end{align*}
Now by the Stirling's approximation and the Taylor series of $\log(1+x)$,
$$ 2\log (K!) = \left(2K + 1\right) \log K - 2 K + \log(2\pi) + \mathcal{O}(K^{-1}) $$
and
$$ K^2 \log\left(1 + K^{-1}\right) = K - \frac{1}{2} + \mathcal{O}(K^{-1}) $$
as $K \to \infty$. Plugging this back to $S_K$, we get
$$ S_K =  \frac{3}{2} - \log \pi + \mathcal{O}(K^{-1}) $$
and the desired identity follows by letting $K\to\infty$.

2nd Solution. We begin by noting that the Taylor expansion of the digamma function
\begin{align*}
\psi(1+z)
&= -\gamma + \sum_{k=1}^{\infty} (-1)^{k-1} \zeta(k+1) z^{k} \\
&= -\gamma + \zeta(2) z - \zeta(3) z^2 + \zeta(4) z^3 - \dots, 
\end{align*}
holds for $|z| < 1$. Then by the Abel's Theorem,
\begin{align*}
S
&= \int_{0}^{1} \sum_{n=1}^{\infty} 2 (\zeta(2n)-1) x^{2n+1} \, \mathrm{d}x \\
&= \int_{0}^{1} x^2 \left( \psi(1+x) - \psi(1-x) - \frac{2x}{1-x^2} \right) \, \mathrm{d}x \\
&= \int_{0}^{1} x^2 \left( \psi(1+x) - \psi(2-x) + \frac{1}{1+x} \right) \, \mathrm{d}x, \tag{1}
\end{align*}
where the identity
$$ \psi(1+z) = \psi(z) + \frac{1}{z} \tag{2} $$
is used in the last step. Then by using the substitution $x\mapsto 1-x$, we get
$$ \int_{0}^{1} x^2 \psi(2-x) \, \mathrm{d}x
= \int_{0}^{1} (1-x)^2 \psi(1+x) \, \mathrm{d}x. $$
Plugging this back to $\text{(1)}$ and performing integration by parts,
\begin{align*}
S
&= \int_{0}^{1} (2x-1) \psi(1+x) \, \mathrm{d}x + \int_{0}^{1} \frac{x^2}{1+x} \, \mathrm{d}x \\
&= -2 \int_{0}^{1} \log\Gamma(1+x) \, \mathrm{d}x + \int_{0}^{1} \frac{x^2}{1+x} \, \mathrm{d}x.
\end{align*}
Now the integrals in the last step can be computed as
$$ \int_{0}^{1} \log\Gamma(1+x) \, \mathrm{d}x = -1 + \frac{1}{2}\log(2\pi)
\qquad \text{and} \qquad
\int_{0}^{1} \frac{x^2}{1+x} \, \mathrm{d}x = -\frac{1}{2} + \log 2. $$
For instance, the first integral can be computed by writing $\log\Gamma(x+1) = \log\Gamma(x) + \log x$ and applying the Euler's reflection formula. For a more detail, check this posting. Finally, plugging these back to $S$ proves the desired identity.
