Antiderivative of $e^{-|x|}$

Question

Splitting the function into the negative and positive parts of the domain of $x$ gives $$ e^{-|x|}= \left\{ \begin{aligned} &e^x, &x ≤ 0 \\ &e^{-x}, &x > 0, \end{aligned} \right. $$ which is trivial to integrate.

Now, the problem is, the answer given in our professor's quiz, and by Wolfram Alpha, is:

$$ \int e^{-|x|}\ dx= \left\{ \begin{aligned} &e^x + c, &x ≤ 0 \\ -&e^{-x} + 2 + c, &x > 0 \\ \end{aligned} \right. $$

I have absolutely no idea where the $+2$ term could come from.

Answer 1

The $+2$ is added so that the derivative is continuous at $x = 0$.

This has the advantage that the derivative at $x = 0$ also exists and is equal to $1$.

Answer 2

If that $2$ was not there, that function would not even be continuous, and so it would not be differentiable.