 6:06 AM
@TheSimpliFire found an interesting integral
Let $a$ and $b$ represent functions of $x$. Then if $$4\int ab\,\mathrm dx = ab(a^2+b^2)-\log |a+b|$$ then WLOG $a=\sqrt x$ and $b=\sqrt{x+1}$
this is interesting because this is the only case that forces $\log |a+b| = \text{arsinh}(x)$.

3 hours later… 9:37 AM
@MrPie How did you prove uniqueness?
We have $4\int_0^xf(t)g(t)\,dt=fg(f^2+g^2)-\log|f+g|+C$ where $C=4H\vert_{t=0}$ and $H'=fg$. Differentiating, we get $4fg=(f^2+g^2)(f'g+fg')+2fg(ff'+gg')-\frac{f'+g'}{f+g}$ and this looks too complicated to solve.

6 hours later… 3:25 PM
@TheSimpliFire i didn't
intuition :P i knew it worked for $a=\sqrt x$ and $b=\sqrt{x+1}$ since i integrated that and i noticed the pattern
then i realised that sinh thing and my gut instinct went yup something's happening here
but yeah, i didn't prove that those are the only values it works for haha :) Let's see what happens if $g=1$ wlog
We have $4\int f\,dx=f(f^2+1)-\log|f+1|+c$ yeah i'd have no idea hahaha
i mean, maybe i might let $f(x)=mx+n$ to generalise?
see where that goes Differentiating, we have $4f=f'(f^2+1)+2f^2f'-\frac1{f+1}=f'(3f^2+1)-\frac{f'}{f+1}$ which a nonlinear ODE
I'll put that in to wolfram
My laptop is so slow today
@MrPie We get solutions! In closed form! 3:33 PM
dammit! Oh wait
I put 1/(f+1) instead of f'/(f+1) :| Even better :P
Ramanujan, ur leaving me no choice but to start believing in the elephant goddess. because clearly this can't be fluked. The family of nontrivial solutions $f(x)$ satisfies $$\frac32(f^2-1)+\log(f+1)=4x+C$$
Actually this equation isn't too hard to solve as it is separable
We might be able to write $f$ in terms of W but it's too much work 3:38 PM
ok i won't be lazy and assume things, let's do this WA can't find an expression with W
But even if there's no closed form, we can show existence. Notice that $f$ is the inverse function of $\frac14\left[\frac32(x^2-1)+\log|x+1|-C\right]$ which is continuous on its domain and monotonically increasing and thus $f$ exists.
So $f=\sqrt x$ and $g=\sqrt{x+1}$ are not unique. i wanna see if i can find some other closed forms I doubt you can
We can do something else though
Plug in $f=\sqrt x$ and find the family of solutions $g$. We know from your calculation that one such solution must be $\sqrt{x+1}$
Again, differentiating yields $4g\sqrt x=(g^2+x)(g/(2\sqrt x)+g'\sqrt x)+2g\sqrt x(1/2+gg')-\frac{1/(2\sqrt x)+g'}{\sqrt x+g'}$
Hm, we really can't solve it ok, i have an idea
let $b=1$ so we have $4\int a\,\mathrm dx = a(a^2+1)-\log |a+1|$ and let $a=mx$. We should get $2x = a^2+1-\frac 1a\log |a+1|$ so that means $$a+\frac 1a\bigg(\frac{\log |a+1|}{a}-\frac{1}{a+1}\bigg)=2$$
would this work? 3:59 PM
@MrPie If $f=mx$ then we get $2mx^2+c=mx(m^2x^2+1)-\log|mx+1|$ which is absurd for any nonzero constant $m$ since the LHS is a polynomial and the RHS is not. Thus $f$ cannot be linear. it's because of that log isn't it Also $a,b$ aren't really used to denote functions since they are more used for constants/integers
@MrPie Yes i was lazy and didn't really want to write $a(x)$ and $b(x)$ over and over lmao I meant use f,g rather than a,b
I've also seen p,q used well, we know there are more than one solution $(a,b)$ wlog
so my theorem is false 4:02 PM
OK if you insist :P well it is, isn't it?
(: @MrPie There are deeper reasons for this. If you're interested you can look at the theory of differential equations
Of course you must have a good knowledge on elementary calculus first okie
give me an integral of a fraction where the numerator is some ordinary polynomial of x, and the denominator is the same thing (but not equal ofc; by same, i mean same in form) You mean like $\int\frac{\sum_{i=0}^na_ix^i}{\sum_{i=0}^nb_ix^i}\,dx$? You'd be better off using the Risch algorithm xd @TheSimpliFire it can prove a hard challenge though: find a solution pair $(f,g)$ such that 4 times the integral blah blah blah 4:06 PM
It is almost always extremely difficult
Many ODEs cannot be solved analytically damn
welp, what is the risch algorithm
maybe let $a_i=b_i=1$ and something like $\frac{x^3+1}{x^4-1}$ idk. It's a computer algorithm to solve integrals computer algorithm?
ew
:P  Risch algorithm
In symbolic computation, the Risch algorithm is an algorithm for indefinite integration. It is used in some computer algebra systems to find antiderivatives. It is named after the American mathematician Robert Henry Risch, a specialist in computer algebra who developed it in 1968. The algorithm transforms the problem of integration into a problem in algebra. It is based on the form of the function being integrated and on methods for integrating rational functions, radicals, logarithms, and exponential functions. Risch called it a decision procedure, because it is a method for deciding whether a... ok im gonna solve $\int \frac{x^3+1}{x^4-1}\,\mathrm dx$ 4:31 PM
@TheSimpliFire know how to solve $\int \frac{w-2}{w^2-14}\,\mathrm dw$?

2 hours later… 6:51 PM
@MrPie Come on, you've done this before :)
Integrals of the form (ax - b)/(cx^2 - d) are nice to evaluate im an idiot Have you had any sleep today :/ no
but i know i can probably let $w=\csc u \sqrt{14}$ @MrPie :/ try to get some (I know I've said this before but) 7:27 PM
@TheSimpliFire how does one do derivatives to two variables at once?
e.g. I want to derive $\frac{b^2-a^2}{b^2+a^2}$ by applying the quotient rule (learnt it now, not at all that different from product rule) but i don't know how to derive this with respect to two values. In this case it's $\dfrac{\partial^2}{\partial a\partial b}$, or equivalently (with some caveats), $\dfrac{\partial^2}{\partial b\partial a}$. ..how does that work And it would be much easier if you write $1-\dfrac{2a^2}{b^2+a^2}$ ok So you differentiate with respect to $a$, whilst treating $b$ as a constant. Then you differentiate with respect to $b$, whilst treating $a$ as a constant. 7:31 PM
order doesn't matter?
(e.g. doesn't matter if i first differentiate with respect to a, or with respect to b?) @MrPie It's a tricky question as you need to know stuff like nth differentiability and continuity
4  I've taken multivariate calculus and am wondering if I can see a specific function where the order of taking the partial derivative matters. I've been told that there are some exceptions where $\dfrac{\partial ^2 f}{\partial x \partial y} \ne \dfrac{\partial ^2 f}{\partial y \partial x}$, so I'...

In your case you have $b^2+a^2$ in the denominator, so let's say that we aren't differentiating around $(a,b)=(0,0)$ Well, I had the integral $\int x\sqrt{\dfrac{1-x^2}{1+x^2}}\,\mathrm dx$. I let $x=\sqrt u$ to give $\int x\sqrt{\dfrac{1-x^2}{1+x^2}}\,\mathrm dx=\dfrac 12\int \sqrt{\dfrac{1-u}{1+u}}\,\mathrm du$ and then I let $u=\frac{b^2-a^2}{b^2+a^2}$ so that $\sqrt{\dfrac{1-u}{1+u}}=\dfrac ab$. Then it doesn't matter
@MrPie What is $u$ a function of? $a$ or $b$? i don't know. I just let $u$ be equal to that, i guess it's... bivariate? $u(a,b)$? It can't be both as your integral is single
One of $a,b$ must be constant 7:36 PM
well actually
there can be more than one antiderivative right? Only if they differ by a constant
Like $\tan^2x$ and $\sec^2x$ or $\pi$ @MrPie So what exactly are you trying to do with it solve the integral :) 7:43 PM
i'm just wondering how i do this part now. i differentiated with respect to b for the first part, since the b's cancel out
I get $2\int \frac{a^3}{(a^2+b^2)^2}\,\mathrm da$
Am I allowed to substitute say $b=\cos \theta$?
this conjugate substitution approach is just weird, i know, but eh it's fun @MrPie Wait what
I think I've lost you wokie, lemme explain
I originally had $$\int x\sqrt{\frac{1-x^2}{1+x^2}}\,\mathrm dx \stackrel{x=\sqrt u}{=}\frac 12\int\sqrt{\frac{1-u}{1+u}}\,\mathrm du$$ since $\mathrm dx = 1/(2\sqrt u)\,\mathrm du$. Note that: $$\cfrac{1-\dfrac{a-b}{a+b}}{1+\dfrac{a-b}{a+b}}=\dfrac ba$$ so if we let $u = \dfrac{b^2-a^2}{b^2+a^2}$ then $\sqrt{\dfrac{1-u}{1+u}}=\dfrac ab.$
Differentiating $u$ with respect to $a$ and $b$ respectively by the quotient rule, we obtain
$$\mathrm du = \dfrac{-2a(b^2+a^2)-2a(b^2-a^2)}{(b^2+a^2)^2}\,\mathrm da = -\dfrac{4ab^2}{(a^2+b^2)^2}\,\mathrm da$$

$$\mathrm du = \dfrac{2b(b^2+a^2)-2b(b^2-a^2)}{(b^2+a^2)^2}\,\mathrm db = \dfrac{4a^2b}{(a^2+b^2)^2}\,\mathrm db$$

It is best to choose $\dfrac{\mathrm du}{\mathrm db}$ because this has a $b$ in the numerator and the integrand upon substitution would be $\dfrac ab$ with a $b$ in the denominator, so these would cancel out. And I just realised I wrote $\mathrm da$ instead of $\mathrm Since$a^3$is merely a constant here, I think I can put it outside the integral.$2a^3\int\dfrac{1}{(a^2+b^2)^2}\,\mathrm db$and now if we let$b=\sqrt{v}$and we get $$2a^3\int\dfrac{1}{(a^2+b^2)^2}\,\mathrm db=a^3\int\dfrac{1}{\sqrt v(a^2+v)^2}\,\mathrm dv$$ 8:14 PM I let$v=w-a^2$so now we have$a^3\int \dfrac{1}{w^2\sqrt{w-a^2}}\,\mathrm dw\$ and im not sure what to do.