Evaluating $\int |x|^3 \; dx $ $$\int |x|^3 \; dx $$
In my module it is suggest to use integration by parts, 
$$ \text{ Set }I = \int (|x|^3 \cdot 1) \; dx = |x|^3 \cdot x - \int \color{red}{\frac {x^3}{|x|^3}3x^2}\cdot x \; dx$$ 
$$ \implies I = |x|^3 \cdot x - \color{red}{3\int |x|^3\;} dx$$
$$ \implies I = \frac 1 4 |x|^3 \cdot x +C $$
I am having trouble understanding the red parts, to be precise the differentiation of $|x|^3 $.
I will also appreciate if somebody  wants to share a different approach (if any).
Thanks,
 A: In retrospect the below is actually not a good answer. Indeed, $x/|x|$ in place of the sign function might have been intended to either avert defining a new function or an attempt to express $\mathrm{sgn}(x)$ purely in terms of elementary functions, but this direction is problematic. The below reasoning works when the region of integration exists strictly on one side of $0$ on the real line, but by its construction it does not work when the region of integration contains $0$; the fundamental theorem of calculus is not valid across singularities.
(More generally, if we remove the appropriate points from the domain to avert the singularities, the FTC only applies on each connected component of the domain individually, and not more than one at a time; the primitive, or antiderivative, associated to each component can be modified by a $+C$ after all.) Technically it is a removable singularity so we can make it work out, but this is an unnecessarily roundabout solution method with no advantage to it.
(Deleting points from the domain and considering $x\mapsto|x|$ a piecewise differentiable function is useful in general for finding an explicit derivative when we have the absolute value composed with other complicated functions, but the same exact caveat applies in the context of integration.)

We can use the chain rule on $f(x)=|x|$ and $g(x)=x^3$, so we reduce to differentiating $|x|$. The derivative does not exist at zero (geometrically, this should be obvious), so we address two possible cases: $x>0$ and $x<0$. In the former, we differentiate as usual and we have $D|x|=1$. In the latter we have $|x|=-x$ so we conclude $D|x|=-1$. (Again this should be obvious, geometrically.) This is the sign function, and it can be written $\mathrm{sgn}(x)=x/|x|$ outside of $x=0$. Applying these facts,
$$\frac{d}{dx}|x|^3=\frac{d}{dx}|x^3|=\mathrm{sgn}(x^3)\cdot \frac{d}{dx}x^3=\frac{x^3}{|x^3|}\cdot3x^2=\frac{x^3}{|x|^3}\cdot3x^2.$$
At least I assume that's what they're getting at.
A: Let $u:x\mapsto|x|^3$. Then $u(x)=x^3$ for $x\gt0$ hence $u'(x)=3x^2$ for $x\gt0$. Likewise $u(x)=-x^3$ for $x\lt0$ hence $u'(x)=-3x^2$ for $x\lt0$. Finally, a direct computation yields $u'(0)=0$. To sum up, $u$ is differentiable everywhere and $u'(x)=\mathrm{sgn}(x)\cdot3x^2$ for every $x$, where $\mathrm{sgn}(x)$ is defined as the sign of $x$ if $x\ne0$ and as anything one wants if $x=0$. 
Alternative formulations of $\mathrm{sgn}(x)$ when $x\ne0$ are $x/|x|$ or $x^3/|x|^3$ or a number of equivalent other ones, but I fail to see their advantage over the canonical expression of $u'$ given above.
In the end, modulo the odd expression of $u'(x)$, the method in this post is a perfectly legitimate application of the integration by parts formula $\int uv'=\left[uv\right]-\int u'v$, to the functions $u$ as above, and $v:x\mapsto x$ with derivative $v':x\mapsto1$. Note that the functions $u$, $v$, $u'$ and $v'$ used in the integration by parts are defined everywhere, the point $0$ included.
A: Recall that $|x| = \sqrt {x^2}$. So $|x|^3 = (\sqrt {x^2}) ^3 = (x^2)^{3/2}$.  Differentiating , you get $$\frac{ 3}{2} (x^2)^{1/2}\cdot 2x = 3x |x|.$$ 
A: Find the primitive of $x^3$ for $x>0$,and  the primitive of $-x^3$ for $x<0$.
Then the primitive you want is : $\frac{x^4}{4}$ for $x>0$, $\frac{-x^4}{4}$ for $x<0$ and for $x=0$
it must be $0$ since a primitive function is continuous.
(But if you have to use integration by parts this solution is not what you want).
A: Let $\displaystyle I= \int|x|^3 dx$
\begin{align*}I =& \int x^2|x| dx \\
=& |x|\int x^2 dx - \int\frac{d}{dx}(|x|)\left(\int x^2 dx \right)dx \\
=&\frac{|x|x^3}{3} - \int\left(\frac{|x|}{x}\cdot\frac{x^3}{3}\right)dx  \text{ since } \frac{d}{dx}(|x|) = \frac{|x|}{x} \\
=&\frac{|x|x^3}{3} - \frac{1}{3}\int|x|x^2dx \\
=&\frac{|x|x^3}{3}  - \frac{I}{3}\end{align*}
That implies,
\begin{align*}I + \frac{I}{3} =& \frac{|x|x^3}{3} \\
\frac{4I}{3}  =& \frac{|x|x^3}{3} \\
4I =& |x|(x^3)\end{align*}
Therefore $\displaystyle I = \frac{|x|x^3}{4}$
A: Given that $|x|=x \operatorname{sgn}(x)$ have: $|x|^3=x^3\operatorname{sgn}(x)^3$. 
Hence we have:
\begin{align*}
\int |x|^3 ~dx & = \int x^3 \operatorname{sgn}(x)^3 ~dx \\
 & = \operatorname{sgn}(x)^3 \int x^3 ~dx \\
 & = \operatorname{sgn}(x)^3 \frac{x^4}{4} \\
 & = \frac{1}{4}x^3 \operatorname{sgn}(x)^3 x \\
 & = \frac{1}{4}|x|^3 x.
\end{align*}
