Question

integral (x + |x|) dx

Answer 1

well you have two functions here, one for \(x>0\) and one for \(x<0\)

Answer 2

\(x>0\implies |x|=x\) so you have \[\int(2x)dx\] whereas \(x<0\implies |x|=-x\) so you have
\[\int 0dx\]

Answer 3

First, let us define a function sign(x) such that:
sign(x) = 1 if x > 0, -1 if < 0 and 0 if x = 0.
Then let us integrate as two separate x. We get:
\[ \frac{1}{2}x^2 + sign(x)\frac{1}{2}x^2 + C\]Because if x <= 0, we get C. If x > 0, we get the integral of 2x, which is x^2.

Answer 4

And @satellite73 's way works fine too :-)

Answer 5

so your "anti derviatve" is a piecewise function,
\[f(x) = \left\{\begin{array}{rcc}
x^2& \text{if} & x \>0 \\
0& \text{if} & x < 0
\end{array}
\right.
\]
although bmp answer is more elegant