How to integrate $|x| \cdot x$ How to integrate this manually?
$$
\int |x|\cdot x ~dx 
$$
My tries so far:
$$
\int |x|\cdot x ~dx = (x^2/2)\cdot|x| - \int (x²/2)\cdot \mathop{\mathrm{sign}}(x) ~dx 
$$
Trying it again, but using sign(x) as first parameter, because sign(x) is not derivable further.
$$
 \int \mathop{\mathrm{sign}}(x)\cdot(x²/2) ~dx =|x|\cdot (x^2/2) - \int |x|\cdot x ~dx 
$$
Great, as nothing would have been done.
Next try, using the signum function
$$
|x|\cdot x = \mathop{\mathrm{sign}}(x)\cdot x^2
$$
$$
 \int \mathop{\mathrm{sign}}(x)\cdot |x| ~dx = x^2-\int|x|\cdot x^2~dx
$$
$$
 \int |x|\cdot x^2 ~dx =x²\cdot \mathop{\mathrm{sign}}(x)\cdot x^2-\int x^2\cdot \mathop{\mathrm{sign}}(x)\cdot 2x ~dx
$$
which seems to be a never ending chain again.
Any ideas?
 A: For $x>0$, this is the integral of $x^2$ which is $\frac{x^3}{3}$. For $x<0$, is the integral of $-x^2$, which is $-\frac{x^3}{3}$. This is just
$\frac{|x^3|}{3}$
A: $$\begin{align*}
\int x|x|\,dx &= \int_c^x t|t|\,dt \\
&= \begin{cases}\begin{cases} \int_c^x t^2\, dt, & c\ge 0 \\ \int_0^x t^2\,dt - \int_c^0 t^2\,dt, & c < 0\end{cases}, & x\ge 0 \\ \begin{cases} \int_x^0 t^2\, dt - \int_0^c t^2\,dt, & c\ge 0 \\ -\int_c^x t^2\,dt, & c < 0\end{cases}, & x< 0\end{cases} \\
&= \begin{cases}\begin{cases} \frac 13(x^3-c^3), & c\ge 0 \\ \frac 13(x^3 +c^3), & c < 0\end{cases}, & x\ge 0 \\ \begin{cases} -\frac 13(x^3+c^3), & c\ge 0 \\ -\frac 13(x^3-c^3), & c < 0\end{cases}, & x< 0\end{cases} \\
&= \frac 13|x|^3 + \text{const}\end{align*}$$
A: I will use the sign function $\text{sgn} (x)$ you attempted to use in your original solution to the problem.
The sign function is defined as
$$\text{sgn} (x) =  \begin{cases}
-1, & x < 0\\
0, & x = 0\\
1 & x > 0.
\end{cases}$$
So we see the sign function is independent of $x$ for all real $x$. Also, since for all real $x$ we have 
$$|x| = \text{sgn}(x) \cdot x,$$
the integral can be rewritten as
$$\int x \cdot |x| \, dx = \int x \cdot (\text{sgn} (x) \cdot x) \, dx = \text{sgn}(x) \int x^2 \, dx.$$
Integrating we have
$$\int x \cdot |x| \,dx = \text{sgn}(x) \cdot \frac{x^3}{3} + C = \frac{x^2}{3} \cdot (\text{sgn}(x) \cdot x) + C = \frac{x^2 |x|}{3} + C.$$
A: $\begin{eqnarray}
\int x|x|\,dx&=&\int x|x|\cdot\frac{x}{x}\,dx\\
&=&\int x^2\cdot\frac{|x|}{x}\,dx
\end{eqnarray}$
Let $u=|x|$. Then $du=\frac{|x|}{x}\,dx$.
So
\begin{eqnarray}
\int x^2\cdot\frac{|x|}{x}\,dx&=&\int u^2\,du\\
&=&\frac{u^3}{3}+c\\
&=&\frac{1}{3}x^2|x|+c
\end{eqnarray}
