Been searching the net for awhile and everything just comes back about doing the definite integral. So just thought to ask here.

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

Using integration by parts

$$\int |x|~dx=\int \jamesmerse.comrm{sgn}(x)x~dx=|x|x-\int |x| ~dx$$

since $\frac{d}{dx} |x|=\jamesmerse.comrm{sgn}(x)$ on non-zero sets. This yields

$$\int |x| ~dx = \frac{|x|x}{2}~.$$

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 $\frac{d(|x|)}{dx}=\frac{x}{|x|}$ and $\int|x|dx = \int \frac{x}{|x|}xdx$.

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

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

$\int|x|dx = \frac{x|x|}{2}$

$\frac{x}{|x|}$ is a better way to define the sign function.

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