Using the integral representation of Bernoulli numbers I obtain formally the integral representation of the double summation $$ \sum_{k=1}^{\infty}\sum_{j=0}^{k}\binom{k}{j}\frac{B_{j+k+1}}{j+k+1}=2\int_0^\infty\frac{t}{e^{2 \pi t}-1}\frac{dt}{t^2+(1+t^2)^2}=0.069591059035995961110566767049... $$ So the alternative form of the question is $$ \int_0^\infty\frac{t}{e^{2 \pi t}-1}\frac{dt}{t^2+(1+t^2)^2}=\frac14+\frac{\ln\phi}{1-2\phi} $$

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}

Numerical calculation in Mathematica gives $$ \sum_{k=1}^{\infty}\sum_{j=0}^{k}\binom{k}{j}\frac{B_{j+k+1}}{j+k+1} =0.069591059035995961110566767049... $$ This quantity is positive, while $\frac{2\,\log\phi}{1-2\phi}=-0.4304089410...$ is negative. So the identity under consideration is wrong.

Using the integral representation of Bernoulli numbers I obtain formally the integral representation of the double summation $$ 2\text{Re}\int_0^\infty\frac{t}{e^{2 \pi t}-1}\frac{1+i t}{t^2-i t+1}dt=0.069591059035995961110566767049... $$

Using the integral representation of Bernoulli numbers I obtain formally the integral representation of the double summation $$ \sum_{k=1}^{\infty}\sum_{j=0}^{k}\binom{k}{j}\frac{B_{j+k+1}}{j+k+1}=2\int_0^\infty\frac{t}{e^{2 \pi t}-1}\frac{dt}{t^2+(1+t^2)^2}=0.069591059035995961110566767049... $$ So the alternative form of the question is $$ \int_0^\infty\frac{t}{e^{2 \pi t}-1}\frac{dt}{t^2+(1+t^2)^2}=\frac14+\frac{\ln\phi}{1-2\phi} $$