Let’s start with a deformation

Area hyperbolic tangent

If we consider the integral


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




Now we substitute





It reminds us of


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





will be


and we’re in the home stretch:


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



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


and then at


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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.



That reminds us of


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


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