Solving $\int e^{|x|} dx$ Solving $\displaystyle\int e^{|x|} dx$.

I just solved a quite similar definite integral, and it wasn't hard, but it was because I could divide the integral in two integrals and get rid of modulus. Problem is I can't do that here, since it's indefinite now. I tried already substitution, parts, series and got nothing. How can I aproach such integral?

Thanks.

Edit: I'd like to obtain Wolfram's answer, not a function defined by parts.
Let us express the antiderivative as a definite integral plus a constant.

$$\int e^{|x|}dx=\int_0^x e^{|x|}dx+C.$$

Then for positive $x$,

$$e^x-1+C$$

and for negative $x$,

$$1-e^{-x}+C.$$

Hence

$$\text{sgn}(x)\left(e^{|x|}-1\right)+C.$$



As we can check, the derivative is $\text{sgn}^2(x)e^{|x|}$ (and $1$ for $x=0$).



Note that WA seemed to strive to avoid the absolute value in the exponent. We can achieve the same effect with 

$$\text{sgn}(x)\frac{(1+\text{sgn}(x))(e^x-1)+(1-\text{sgn}(x))(e^{-x}-1)}{2}
=\frac{(\text{sgn}(x)+1)(e^x-1)-(\text{sgn}(x)-1)(e^x-1)e^{-x}}2
=e^{-x}\frac{\text{sgn}(x)(e^x-1)^2+e^{2x}-e^x-e^x-1}2.$$

This is ugly.
Write $e^{|x|}=e^{\operatorname{sgn}(x) x}$. Then

$$\int e^{|x|} dx = \int e^{\operatorname{sgn}(x)x} dx = \frac{e^{\operatorname{sgn}(x)x}}{\operatorname{sgn}(x)} + C = \operatorname{sgn}(x)e^{|x|} + C$$

edit: And well, this is broken.