How to evaluate $\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{n+1}$ directly?
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.