Let’s start with a deformation

Area hyperbolic tangent

If we consider the integral

    \[\int\frac{\mathrm{d}x}{x^2+4x}\]

it would be nice to know where it came from. It can’t be a derivative of \ln(x^2+4x) because we miss 2x+4 in the numerator. Could it originate from another known function? Let’s add 4 in the denominator and let’s immediately subtract 4 so nothing changes. We get

    \[\int\frac{\mathrm{d}x}{x^2+4x+4-4}\]

or

    \[\int\frac{\mathrm{d}x}{(x+2)^2-4}\]

Now we substitute

    \[u=x+2\]

    \[\mathrm{d}u=\mathrm{d}x\]

so

    \[\int\frac{\mathrm{d}u}{u^2-4}\]

It reminds us of

    \[\artanh'(x)=\frac{1}{1-x^2}\]

Let’s make u four times larger to get rid of the -4.
Substitution again:

    \[u=2v\]

    \[\mathrm{d}u=2\mathrm{d}v\]

so

    \[\int\frac{\mathrm{d}v}{(2v)^2-4}=\int\frac{2\mathrm{d}v}{4v^2-4}\]

will be

    \[\tfrac{1}{2}\int\frac{\mathrm{d}v}{v^2-1}=-\tfrac{1}{2}\int\frac{\mathrm{d}v}{1-v^2}\]

and we’re in the home stretch:

    \[-\tfrac{1}{2}\artanh(v)+C\]

We revert to x with v=\tfrac{1}{2}u=\frac{x+2}{2} and so

    \[F(x)=-\tfrac{1}{2}\artanh\left(\frac{x+2}{2}\right)+C\]

Arctangent

Manhandling a circle will leave you with an ellipse. Manhandling an ellipse will leave you with a parabola. Manhandling a parabola will leave you with a hyperbola. They are all all the result of an intersection with a cone.
So let’s look at

    \[\frac{1}{x^2+4x}\]

and then at

    \[\frac{1}{x^2+4x+8}\]

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That 8 gives us two points with asymptotes from here to infinity: -4 and 0. But tilt the hyperbola far enough in 3D and you will get close to your your original x^2+4x. Would that mean tilting in the cone as well? That would mean we would go from a hyperbolic function \artanh to some trigonometric function of a circle or ellipse, like \arcsin, \arccos or \arctan.
Well,

    \[\int\frac{\mathrm{d}x}{x^2+4x+8}=\int\frac{\mathrm{d}x}{x^2+4x+4+4}\]

    \[\int\frac{\mathrm{d}x}{(x+2)^2+4}\]

That reminds us of

    \[\arctan'(x)=\frac{1}{x^2+1}\]

follow the same steps and without further ado we would end up with

    \[\tfrac{1}{2}\int\frac{\mathrm{d}v}{v^2+1}=\tfrac{1}{2}\arctan\left(\frac{x+2}{2}\right)+C\]

From \artanh to \arctan. From hyperbola via parabola to ellipse and circle.

Well?