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?
 A: 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. 
A: Long method
One possible method is the following. Using 
\begin{align}
\sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{(-1)^n \, x^n}{4^n} &= \frac{1}{\sqrt{1 + x}} \\
\gamma(s,x) &= \int_{0}^{x} e^{-t} \, t^{s-1} \, dt
\end{align}
then:
\begin{align}
I &= \int_{-2}^{0} \frac{x \, dx}{\sqrt{e^{x} + (x+2)^{2}}} \\
&= \int_{-2}^{0} \frac{x \, e^{-x/2} \, dx}{\sqrt{1 + \left(\frac{(x+2)^2}{e^{x}}\right)}} \\
&= \sum_{n} \binom{2n}{n} \frac{(-1)^{n}}{4^{n}} \, \int_{-2}^{0} e^{-(n+1/2) x} \, x \, (x+2)^{2n} \, dx \\
&= \sum_{n} \binom{2n}{n} \frac{(-1)^{n}}{4^n} \, e^{2n+1} \, \int_{0}^{2} e^{-(n+1/2) t} \, (t-2) \, t^{2n} \, dt   \hspace{5mm} t \to x + 2 \\
&= \sum_{n} \binom{2n}{n} \, \frac{(-1)^{n}}{4^{n}} \, e^{2n+1} \, \left[ \left(\frac{2}{2n+1}\right)^{2n+2} \, \gamma(2n+2, 2n+1) - 2 \, \left(\frac{2}{2n+1}\right)^{2n+1} \, \gamma(2n+1, 2n+1) \right]   \\
&= 4 \, \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{(-1)^{n} \, e^{2n+1}}{(2n+1)^{2n+2}} \, \left[ \gamma(2n+2, 2n+1) - (2n+1) \, \gamma(2n+1, 2n+1) \right] \\
&= - 4 \, \sum_{n} \binom{2n}{n} \, \frac{(-1)^{n} \, e^{2n+1}}{(2n+1)^{2n+2}} \, (2n+1)^{2n+1} \, e^{-(2n+1)} \\
&= -4 \, \sum_{n=0}^{\infty} \binom{2n}{n} \, \frac{(-1)^{n}}{(2n+1)} \\
&= - \, \sum_{n} \binom{2n}{n} \frac{(-1)^{n}}{4^{n}} \, \int_{0}^{2} y^{2n} \, dy \\
&= -2 \, \int_{0}^{2} \frac{dy}{\sqrt{1+y^2}} \\ 
&= -2 \, \sinh^{-1}(2).
\end{align}
Short method
\begin{align}
I &= \int_{-2}^{0} \frac{x \, dx}{\sqrt{e^{x} + (x+2)^{2}}} \\
&= -2 \int_{-2}^{0} \frac{- (x/2) \, e^{-x/2} \, dx}{\sqrt{1 + ((x+2) \, e^{-x/2})^2}} \\
\end{align}
Let $u = (x+2) \, e^{-x/2}$, $du = -(x/2) \, e^{-x/2} \, dx$, then
\begin{align}
I &= -2 \int_{-2}^{0} \frac{du}{\sqrt{1+u^2}} = -2 \left[\sinh^{-1}(u)\right]_{-2}^{0} \\
&= -2 \, \sinh^{-1}(2).
\end{align}
It is of interest to note that
$$\int_{-2}^{\infty} \frac{x \, dx}{\sqrt{e^{x} + (x+2)^{2}}} = 0.$$
