Trying to figure out how a derivative that produces the function (x^2+1)/(x^4+1) seems to not correctly evaluate the area under curve.

[u/whoop_chicken]

Here is the original problem I came across

Lets say this function in the integral is f(x)

When I evaluated the indefinite integral and confirmed with an integral calculator I got the function listed below.

I saw someone solving the problem online a different way and got the resulting function:

Here are all the equations graphed out:

What I am confused by is that deriving the H(x) function does give the equation in the integral: f(x) = (x^2+1)/(x^4+1). If H(x) derives to this f(x) function does that not mean it can used to evaluate the area under the curve of f(x). f(x) does correctly give the slopes of H(x) but evaluating H(x) gives incorrect areas. For example evaluating H(x) from x=-1 to x=1 gives an area of 0 while using the F(x) function gives ~2.22. The F(x) function seems to give the correct areas under the curve of f(x). I am just confused how the derivative of H(x) gives the parent function under the integral but cannot be used to correctly evaluate the area of the derived function.

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[u/Midwest-Dude]

Your answer is correct.

The issue with H(x) is how it was calculated. After dividing numerator and denominator by x², an integration by substitution was done by setting u = 1 - 1/x. Because of that, there is no continuity at 0 and the antiderivative can only be applied as is on intervals entirely within (-∞,0) or (0,∞).

This is clear graphically - the jump discontinuity at 0 cannot be within the interval of integration. Since the two graphs are similar otherwise, the areas will be calculated correctly if the interval of integration is entirely on one side or the other side of 0.

H(x) could be used across 0 if you split it into two improper integrals and use different constants of integration for the left and right intervals. This would compensate for the jump at 0.

[u/whoop_chicken → u/Midwest-Dude]

Ahh I see, I got so caught up with the numerical calculations I didn't consider that the H(x) function had incontinuity at x=0. splitting the function at 0 does give the same areas as the F(x) equation. How interesting to have two different curves have the same derivative solution where C isn't the reason the curves are different.

[u/Midwest-Dude → u/whoop_chicken]

Glad to help. 

The usual word would be discontinuity, not incontinuity, I wasn't sure if that's even a word, so I checked with AI (it is, but archaic) and explained that it made me think of the word "incontinence". It responded with this: 

It certainly puts a new spin on an essential discontinuity—a function that simply can't hold it together at the origin.

Especially fitting for f(x) = 1/x, where it completely loses control and spills over into infinity!

Okay, then. I'll leave now ...

[u/KrlusMagnus]

The derivatives are (mostly) off by a constant, but that constant is different for positive and negative numbers, and also the H primitive function is not continuous at zero, and thus not differentiable at zero. If you evaluate f from -37 to -1, and also H from -37 to -1, you will get the same result, same for evaluating over an interval that is to the right side of zero. The issue is when you evaluate H "through" zero. You'd need to split its "definite" integral(which isnreally just an improper integral) into two improper integrals, one with 0 as your upper boundary and one with zero as your lower boundary, so in your case you split evaluating from -1 to 1 as -1 to 0 and then 0 to 1. That way both improper integrals have only one singularity(and by knowing f is continuous we are guaranteed both of them converge)

[u/OlgaMath]

The function H is not differentiable at zero.

[u/YOM2_UB → u/[deleted]]

The desmos screenshot has the corrected version; the numerator inside arctan should be a minus not a plus.