Indefinite integral of absolute value When I looked up about absolute value on Wikipedia, I found that the antiderivative of $|x|$ is $\frac12 x|x|+C$. I am able to find the derivative of $|x|$ by treating the function as $\sqrt{x^2}$, but I am not able to integrate it.
When I put $\int_{-4}^{-1}|x|\,dx$ into Symbolab, the online calculator did not break the integral into piecewise function but calculate the indefinite integral first before using $F(b) - F(a)$. When I view the steps it used, it said:
If $\int f(x)\,dx = F(x)$
then $$\int \sqrt{(f(x))^2)}\,dx = \frac{\sqrt{f(x)^2}}{f(x)}$$ multiplied to $F(x)$
which becomes $\frac{\sqrt{x^2}}{x}$ multiplied to $\int x\,dx$
 A: The function ${\rm abs}$ is continuous on all of ${\mathbb R}$, hence should have primitives $F$ defined on all of ${\mathbb R}$. Given that ${\rm abs}$ is "special" at $x=0$ we should look for the primitive obtained by integrating from $0$ to $x$. In this way we obtain
$$F(x)=\int_0^x |t|\>dt=\int_0^x t\>dt={x^2\over2}\qquad(x\geq0)$$
and
$$F(x)=\int_0^x |t|\>dt=\int_0^x (-t)\>dt=-{x^2\over2}\qquad(x\leq0)\ .$$
The two partial results can be condensed into the single formula
$$F(x)={x\>|x|\over 2}\qquad(-\infty<x<\infty)\ ,$$
and adding an arbitrary constant $C$ gives the general primitive of ${\rm abs}$.
A: Here is my derivation for this,
$$\int |x|dx= \int \sqrt {x^2} dx = I$$
By integration by parts we know that, 
$$ \int u(x)v(x)\,dx=u(x)\int v(x) \ dx-\int u'(x) \left [ \int v(x) \ dx \right ]dx$$
This is just a extension of the product rule in diffrentiation. Check out the wiki page.
So taking $u(x) = |x|$ and $v(x) =1$,
$$I = \int |x| \times 1 \ \  dx = |x| \int 1\ dx - \int \left [ \frac{d(|x|) }{dx}\int 1 \ dx \right ]dx $$ 
Now we can differentiate the absolute value of $x$ using chain rule,
$$\frac{d(|x|) }{dx} = \frac{d(\sqrt {x^2}) }{dx} = \frac{1}{2\sqrt{x^2}} (2x)= \frac{x}{\sqrt{x^2}} $$
Trivially we can say $ \int 1 \ dx = x$.
Substituting this in $I$,
$$I = |x|x \ - \ \int\frac{x}{\sqrt{x^2}} \ x \ dx$$
$$I = |x|x \ - \ \int\frac{x^2}{\sqrt{x^2}} \ dx$$
Because both $x^2$ and $\sqrt{x^2}$ are positive, we can rewrite this as,
$$I = |x|x \ - \ \int{\sqrt{x^2}} \ dx =|x|x \ - I $$
So as we have the same integral in the RHS, we take it to the LHS.
$$2I = |x|x$$
So we can conclude, 
$$I(x) = \frac{x|x|}{2}$$
Try this out for yourself the area under the $|x|$ curve from $x=a$ to $x=b$ can be expressed as $I(b) - I(a)$.
The antiderivative of $|x|$ is a function $g(x)$ such that $g'(x) = |x|$. Note that for any value of $C$, $g(x) + C$ can also be such an antiderivative. So we add a Constant of integration.

SIDE NOTE : 
In case the integration by parts formula I gave above is confusing,
  consider
$$\frac{d(a(x) \ b(x)) }{dx} = a'(x) \ b(x) + a(x) \ b'(x)$$
Integrating both sides,
$$\int \frac{d(a(x) \ b(x)) }{dx} = a(x) \ b(x) = \int a'(x) \ b(x) + \int a(x) \ b'(x)$$
If we substitute $a(x) = u(x)$ and $b'(x) = v(x)$, so  $b(x) = \int v(x)$ and $a'(x) = u'(x)$
$$u(x)\int v(x) = \int u'(x) \left [ \int v(x) \ dx \right ]dx + \int u(x) \ v(x)$$
Thus we get the formula for integration by parts shown above.

A: enter image description here
∫|x|dx=∫〖√(x^2 ) dx〗=
=x√(x^2 )-∫〖x/√(x^2 ) dx〗
Then:
∫〖√(x^2 ) dx〗=x√(x^2 )-∫〖√(x^2 ) dx〗
∫〖√(x^2 ) dx〗=x√(x^2 )-∫〖√(x^2 ) dx〗
Then:
2∫〖√(x^2 ) dx〗=x√(x^2 )=x|x|
Finally we will have:
  ∫〖√(x^2 ) dx〗=x|x|/2

I can not put the formula correctly, If you guide how to do the job I will provide the answer in a better view
A: $\int_{-4}^{-1}|x|~dx$
$=[x|x|]_{-4}^{-1}-\int_{-4}^{-1}x~d(|x|)$
$=15-\int_{-4}^{-1}x\times\dfrac{|x|}{x}dx$
$=15-\int_{-4}^{-1}|x|~dx$
$\therefore2\int_{-4}^{-1}|x|~dx=15$
$\int_{-4}^{-1}|x|~dx=\dfrac{15}{2}$
