Bernoulli sum meets golden number

Using the integral representation of Bernoulli numbers I obtain formally the integral representation of the double summation $$ \sum_{k=1}^{\infty}\sum_{j=0}^{k}\binom{k}{j}\frac{B_{j+k+1}}{j+k+1}=2\cdot\int_0^\infty\frac{t}{e^{2 \pi t}-1}\frac{dt}{t^2+(1+t^2)^2}=0.069591059035995961110566767049... $$ So the alternative form of the question is $$ \int_0^\infty\frac{t}{e^{2 \pi t}-1}\frac{dt}{t^2+(1+t^2)^2}=\frac14+\frac{\ln\phi}{1-2\phi} $$ $\it{Proof}.$ Since $$ \frac{1}{t^2+(1+t^2)^2}=\frac{1}{\sqrt{5}}\left(\frac1{t^2+1/\phi^2}-\frac1{t^2+\phi^2}\right), $$ and according to Binet's second integral representation for the digamma function $\psi$ $$ \psi(\phi)=-\frac1{2\phi}+\ln\phi-2\int_0^\infty\frac{t}{e^{2 \pi t}-1}\frac{dt}{t^2+\phi^2}, $$ $$ \psi(1/\phi)=-\frac{\phi}{2}-\ln\phi-2\int_0^\infty\frac{t}{e^{2 \pi t}-1}\frac{dt}{t^2+1/\phi^2}, $$ and $$ \psi(\phi)-\psi(1/\phi)=\frac1{\phi-1} $$ one has \begin{align} &\int_0^\infty\frac{t}{e^{2 \pi t}-1}\frac{dt}{t^2+(1+t^2)^2}\\ &=\frac1{2\sqrt5}\left(\psi(\phi)-\psi(1/\phi)+\frac1{2\phi}-\frac{\phi}2-2\ln\phi\right)\\ &=\frac14+\frac{\ln\phi}{1-2\phi}. \end{align}

OK, in fact this is easy: simply first prove that $\sum_{j=k}^{2k}\binom{k}{j-k}\frac{B_{j+1}}{j+1}=(-1)^k\binom{2k}{k}\frac{1}{4k+2}$, the rest is immediate.

Edit. Henri, I'm editing. If you don't agree, please delete it. We need to prove that $$\sum_{j=k}^{2k}\binom{k}{j-k}\frac{B_{j+1}}{j+1}=\frac{(-1)^{k-1}}{\binom{2k}{k}(4k+2)}.$$ We also need to prove that $$\sum_{k\geq0}\frac{(-1)^{k-1}}{\binom{2k}{k}(4k+2)}=\frac{2\log\phi}{1-2\phi}.$$

Here is a short proof for the 2nd formula in T. Amdeberhan's edit to H. Cohen' post.

The key point is to observe that \begin{equation*} \frac{1}{\binom{2k}{k}(4k+2)} = \frac12 B(k+1,k+1), \end{equation*} where $B(\cdot,\cdot)$ is the beta function. With this observation we obtain \begin{equation*} \frac12\sum_{k\ge 0}(-1)^{k-1}B(k+1,k+1) = \frac12\int_0^1 \Bigl(\sum_{k\ge 0}(-1)^{k-1}t^k(1-t)^k\Bigr)dt = \frac12\int_0^1 \frac{-1}{1+t-t^2}dt \end{equation*} which is easily verified to equal $\frac{2\log\phi}{1-2\phi}$.