Calculate $ \int_{-2}^{0} \frac{x}{\sqrt{e^x+(x+2)^2}} dx $

It is a tricky one! $$\mathcal{J}=\int_{-2}^{0}\frac{x}{\sqrt{e^x+(x+2)^2}}\,dx=\int_{-2}^{0}\frac{x e^{-x/2}}{\sqrt{1+\left[(x+2)e^{-x/2}\right]^2}}\,dx $$ and you may notice that up to a multiplicative constant the numerator $xe^{-x/2}$ is the derivative of $(x+2) e^{-x/2}$. Additionally the function $(x+2)e^{-x/2}$ is increasing on $(-2,0)$, going from $0$ to $2$.
By substituting $(x+2)e^{-x/2}=z$ we get $$ \mathcal{J}= -2\int_{0}^{2}\frac{dz}{\sqrt{1+z^2}} = -2\,\text{arcsinh}(2) =-6\log\varphi$$ as claimed in the comments. As a folklore note , my (pretty old) version of Mathematica is unable to recognize such closed form for $\mathcal{J}$. A point for humans vs machines, I guess.