Title says it all. Is there a closed form solution for the indefinite integral $int |x| dx$ ?

Using integration by parts

$$int |x|~dx=int hedailysplash.tvrmsgn(x)x~dx=|x|x-int |x| ~dx$$

since $fracddx |x|= hedailysplash.tvrmsgn(x)$ on non-zero sets. This yields

$$int |x| ~dx = fracx2~.$$

You are looking for a function $f(x)$ so that $$int_a^b |x|dx=f(b)-f(a).$$ This is what is meant by $int |x|dx$. I propose that $f(x)=x|x|/2$ is such a function. Let us test it. If both $a$ and $b$ are both positive, then $$int_a^b |x|dx=int_a^b x,dx=b^2/2-a^2/2=b|b|/2-a|a|/2=f(b)-f(a).$$If $a$ and $b$ are both negative, then $$int_a^b |x|dx=-int_a^b x,dx=-b^2/2-(-a^2/2)=b|b|/2-a|a|/2=f(b)-f(a).$$Finally, if $a and $b>0$, we get$$int_a^b |x|dx=-int_a^0 x,dx+int_0^b x,dx=b^2/2+a^2/2=b|b|/2-a|a|/2=f(b)-f(a).$$Of course, we could have $b and $a>0$, but then we could switch the limits, and this reduces to the third case.

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Thus, $f(x)=x|x|/2$ is an indefinite integral, or antiderivative of $|x|$.

You can use $frac)dx=fracx$ and $int|x|dx = int fracxxdx$.

$int|x|dx = int xd(|x|)$, using integration by parts $int|x|dx = x|x| - int|x|dx $

$2int|x|dx = x|x|$

$int|x|dx = fracx2$

$fracxx$ is a better way to define the sign function.

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