# Is this really true?

Any insights on this monstrous integral?

$\int\frac{x}{\sqrt{x^4+10x^2-96x-71}} \, dx = -\frac18 \ln((x^6+15x^4-80x^3+27x^2-528x+781)\sqrt{x^4+10x^2-96x-71}-(x^8+20x^6-128x^5+54x^4-1408x^3+3124x^2+10001))+constant$ Note by Inquisitor Math
8 months, 1 week ago

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Hi, chris! I’m so happy to see you again! There is not much context for this thing, but this integral is said to be found by ‘Manuel Bronstein’ who tried to find an elementary function that has an elementary derivative.(tho this thing is not easy to find. how did he come up with this thing??)

- 8 months ago

OK, so Manuel Bronstein was heavily involved in the development of computational symbolic integration. This - and the fact Alpha struggles with the integral - suggests it's a hard problem!!

The best I can come up with is working backwards: you might observe that integrals of the form $\int \frac{x}{\sqrt{P(x)}} dx$

have a form something like $a \log \left(Q(x) \sqrt{P(x)}-R(x) \right)+C$

where $P,Q,R$ are monic polynomials, $a$ is a constant and $C$ is an arbitrary constant. Let's assume $P(x)$ is not a perfect square.

Differentiating this, we get $a \frac{Q'(x) \sqrt{P(x)} + \frac{P'(x) Q(x)}{2\sqrt{P(x)}} - R'(x)}{Q(x) \sqrt{P(x)}-R(x)} = \frac{x}{\sqrt{P(x)}}$

Multiplying up, $a \left(P(x) Q'(x) + \frac12 P'(x) Q(x) - R'(x) \sqrt{P(x)} \right) = xQ(x) \sqrt{P(x)}-xR(x)$

We want this to be an identity that holds for all $x$. This means we can equate coefficients, and also the terms that are multiples of $\sqrt{P(x)}$ (remember we said $P$ wasn't a square).

So $a \left(P(x) Q'(x) + \frac12 P'(x) Q(x)\right) = -xR(x)$

and $-a R'(x) = xQ(x)$

(it's easy to verify these hold for the example). This certainly looks neater. We can see from here how the degrees of the polynomials are related: $\deg(P) + \deg(Q)-1=\deg(R)+1,\;\;\;\deg(R)-1=\deg(Q)+1$

so in fact we have to have $\deg(P)=4$. OK, that was a surprise!

Also, we get $a=\frac{-1}{\deg(R)}$ (the two equations both give the same result).

The sensible thing to do here seems to be to pick a polynomial $Q$, use the second equation to get $R$ (up to a constant of integration), then equate coefficients in the first equation. The trickier part (and perhaps why the polynomials in the example have strange looking coefficients) would be getting all of the polynomials to have integer coefficients.

That's as far as I can get at the moment - I'd be interested in your thoughts. Do you think this was the approach? The fact that $\deg(P)$ has to be $4$ is making me reconsider. It'd be worth trying to solve those two equations to see if there are other examples.

- 7 months, 4 weeks ago

I just tried this out. If we set $Q(x)=x^2+3x+4$ (this was arbitrary, though I chose a multiple of $3$ for the $x$ term to make $R$ nicer), we get $a=-\frac14$, $R(x)=x^4 + 4x^3 + 8x^2 - \frac{13}{2}$ and $P(x)=x^4 + 2x^3 + 3x^2 - 12x + 8$

and you can check that $\int \frac{x}{\sqrt{x^4 + 2x^3 + 3x^2 - 12x + 8}}=-\frac14 \log \left( \left(x^2+3x+4 \right) \sqrt{x^4 + 2x^3 + 3x^2 - 12x + 8} - \left(x^4 + 4x^3 + 8x^2 - \frac{13}{2} \right) \right)+C$

Again, Wolfram|Alpha couldn't do the integral, but can do the differentiation.

- 7 months, 4 weeks ago

Hah, that one looks nasty too! Kudos!

- 7 months, 4 weeks ago

At first glance, I thought this involved some crazy substitution, but this is a nice approach! I will ponder more about this and see where this goes! (By the way have you looked at the other link too? That one is a question posed by me.)

- 7 months, 4 weeks ago

There still might be a way involving substitution, but again, if Alpha can't do it, I don't think I will be able to! (By the way, have you seen the form that Alpha comes out with for the original one? Not nice.)

I wonder what the full set of solutions to this looks like. I mean, $P$ has to be a quartic, but I don't think it can be an arbitrary one.

- 7 months, 4 weeks ago

I did some simple calculations and here are the following results: let the degree of polynomial $R$ be $n$. We choose some nice polynomial (degree 4) for $P$ such that the following has a nice polynomial solution. $2PQ^{\prime\prime\prime}+5P^{\prime}Q^{\prime\prime}+(4P ^{\prime\prime}-2n^2x^2)Q^{\prime}+(P^{\prime\prime\prime} -6n^2x)Q=0$ (IS THIS USEFUL?)

After solving for $Q$ we finally get $R$. (Slightly different from your approach, notice that I am starting from P, then to Q, then to R. However this doesn’t seem to yield any meaningful results since a. we still do not know what P should be b.we do not know the value of n.) In fact the examples above (which chose the right $P$) fits the bill!

- 6 months, 2 weeks ago

Just in case you are interested, for the example you tried out, if we set $n=4, P(x)=x^4+2x^3+3x^2-12x+8$ the eq is $2 (x^4+2x^3+3x^2-12x+8)Q^{\prime\prime\prime}+5(4x^3+6x^2+6x-12)Q^{\prime\prime}+(16x^2+48x+24)Q^{\prime}+(12-72x)Q=0$ and $Q(x)=x^2+3x+4$ is indeed a solution!

- 6 months, 2 weeks ago

Hi again! I tried your way of going Q to R to P (P to R) and I have some questions. A)How did you determine the constant of R (after setting Q as a degree 2 polynomial?) B) Did you ‘just’ try out a degree 2 polynomial to make things simple?

This is just my opinion but for degQ=2 I think this may be generalized. (Have you tried it?)

- 5 months, 1 week ago

$\frac1{a^2}x \int xQ-\frac12P^{\prime}Q-PQ^ {\prime}=0$

As a side note this is a result from wolfram. I’m no expert at these stuff... so I hope you can check it out and see if that can be of any help.

I’ve updated some questions on these as well. LINK

- 5 months, 1 week ago

Hi - I need to remind myself a bit of what I did on this! I'll have a look and get back to you over the weekend.

- 5 months ago

Thank you very much! I think the most helpful info here would be the link from wolfram.

Idk if I am asking too much of you... Recently I’m struggling on this problem... I would be greatful if you check it now if you can. LINK1

- 5 months ago

Any motivation,backstory behind this? I have a strong feeling that this is not arbitrary.

- 8 months ago

Hi again - you asked in a solution comment the other day for some help on this one but I'm not sure how much help I can be. Where did you find this question? It might be useful to start there.

Just some observations, though:

1) Yes, it's really true. Wolfram|Alpha couldn't cope with the integral itself, but differentiating the result does give the right integrand

2) The roots of the polynomial $x^4+10x^2-96x-71=0$ are fairly "nice" (the only surds needed are $\sqrt2$ and $\sqrt3$) which probably isn't a coincidence

3) That would suggest trying to find integrals of similar functions, but I haven't managed to get Alpha to do this

4) This might involve contour integration, though I'm not sure how it'd work for indefinite integrals

- 8 months ago