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

Factor out $e^x$ in the denominator. Once you take the square root, you get $e^{x/2}$ in the denominator. Then, make the substitution:

Let $(x+2)e^{-x/2} = \tan \theta$

$-\dfrac{x}{2}e^{-x/2}dx = \sec^2 \theta d\theta$

At $x=-2$, $\tan \theta = 0$

At $x=0$, $\tan \theta = 2$

So, your integral becomes:

$$\int_{-2}^0 \dfrac{x}{\sqrt{e^x+(x+2)^2}}dx = -2\int_0^{\arctan 2} \sec \theta d\theta = -2\ln(\sqrt{5}+2)$$

Mathematica could not solve this as written, $$ I=\int_{-2}^0 \frac{x}{\sqrt{e^x+(x+2)^2}}dx $$ I introduced a parameter $a$ $$ I(a)=\int_{-2}^0 \frac{x}{\sqrt{a e^x+(x+2)^2}}dx $$ took a Mellin transform with respect to $a$ $$ \mathcal{M}_a[I(a)](s)= \Gamma(s)\Gamma\left(\frac{1}{2}-s\right)\int_{-2}^0 \frac{x \left(\frac{e^x}{(2+x)^2}\right)^{-s}}{\sqrt{\pi}\sqrt{(x+2)^2}}dx $$ Mathematica can solve this $$ \mathcal{M}_a[I(a)](s)= \frac{-4^s\Gamma(s)\Gamma\left(\frac{1}{2}-s\right)}{\sqrt{\pi}s} $$ and the inverse Mellin transform gives \begin{equation} I(a) = -2 \text{arcsinh}\left(\frac{2}{\sqrt{a}}\right) \end{equation} which for $a=1$ checks out numerically as around $I \approx -2.88727$