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

First I tried the substitution $t=x+2$ and obtained $$\int\limits_{0}^{2} \frac{t-2}{\sqrt{e^{t-2}+t^2}}dt$$ and than I thought to write it as $$\int\limits_{0}^{2} (t-2)\frac{1}{\sqrt{e^{t-2}+t^2}}dt$$ and use the fact that $$2(\sqrt{e^{t-2}+t^2})'=\frac{1}{\sqrt{e^{t-2}+t^2}} \cdot(e^{t-2}+2t)$$ Using integration by parts we get that we have to calculate (excluding some terms we know) $$\int\limits_{0}^{2} \sqrt{e^{t-2}+t^2}\cdot \frac{6e^{t-2}-2te^{t-2}+8}{(e^{t-2}+2t)^2}dt$$ which is uglier then the initial problem. Do you have any idea how to solve the problem?

Hint: Let $y = (x+2)e^{-x/2}.$ You should get $I=-2\sinh^{-1}{2}$.

$\displaystyle y = (x+2)e^{-x/2}  \implies dy = -\frac{x}{2}e^{-x/2}\,dx$
and $\displaystyle \frac{x\,dx}{\sqrt{e^x+(x+2)^2}} = \frac{xe^{-x/2}\,dx}{\sqrt{1+(x+2)^2e^{-x}}} = \frac{-2\,dy}{\sqrt{1+y^2}}.$
Therefore we have  $\displaystyle I = \int_{-2}^{0}\frac{x\,dx}{\sqrt{e^x+(x+2)^2}} = -2\int_{0}^{2}\frac{dy}{\sqrt{1+y^2}} = -2\sinh^{-1}(2)$, as claimed. 
