Anonymous at Fri, 22 Mar 2024 11:14:23 UTC No. 16090886

well first you have annals of solved integrals
second you have the trick with the series
third you have wolfgram
fourth you have to rely on geniuses like on mathoverflow

Anonymous at Fri, 22 Mar 2024 12:37:11 UTC No. 16090950

>>16090872
If you have to calculate the integral of the reciprocal of some polynomial, if suffices to find the inverse of the polynomial in the respective formal power series ring. E.g. say we have [math]2+X^2\in \mathbb{Q}[X][/math], then its inverse [math](2+X^2)^{-1}=: \sum_{k\in\mathbb{N}_0} b_kX^k=: p\in \mathbb{Q}[\![X]\!][/math] can be calculated by setting [math](2+X^2)p=1_{\mathbb{Q}[\![X]\!]}[/math] and inductively solving that infinite system of linear equations for the coefficients of p. Then, by the linearity of the integral, you could turn the integral into a series that'll hopefully converge. Not sure how useful this is, though.

Anonymous at Fri, 22 Mar 2024 12:49:53 UTC No. 16090961

>>16090950
Lmao stfu nerd

Anonymous at Fri, 22 Mar 2024 14:55:12 UTC No. 16091097

>>16090872
>How can I solve picrel?
You mean the integral of y?

Just use x = cosh theta, sqrt{x^2-1} = sinh theta and dx = sinh theta, so your integral is just theta = cosh^{-1} x.

Anonymous at Fri, 22 Mar 2024 15:44:15 UTC No. 16091174

>>16090872
Most integrals can't be computed exactly, instead you have to find either numerical methods or employ an approximation scheme, but in general you'll need to have the domain of integration specified, for example:
[eqn] \int^1_0 \frac{dt~e^{xt}}{1+t^2} [/eqn] This isn't easy to integrate (I'm not even sure if it's possible to integrate). But notice that we're integrating with respect to [math] t [/math] and not [math] x [/math], so we can treat [math] x [/math] as a constant of sorts and ask what the value of that integral is as [math] x \to 0 [/math].

So, lets expand [math] e^{\alpha} ~ \text{around} ~ \alpha = 0 [/math]:
[eqn] e^{\alpha} \sim 1 + \alpha + \frac{1}{2}\alpha^2 + \cdots [/eqn] Put [math] \alpha = xt [/math]:
[eqn] e^{xt} \sim 1 + xt + \frac{1}{2} x^2 \alpha^2 [/eqn] Which leads to an integral like:
[eqn] \int^1_0dt ~ \left( \frac{1}{1+t^2} + \frac{xt}{1+t^2} + \frac{x^2t^2}{1+t^2} \cdots \right) [/eqn]
Now we exploit the linearity of the integral of three (hopefully easier) integrals, lets only look at the first two terms:
[eqn] \int^1 _0 \frac{dt}{1+t^2} = \frac{\pi}{4} \\ x\int^1_0 \frac{t}{1+t^2} = \ \frac{x\ln(2)}{2}[/eqn]So! After all of this we see that:

[eqn] \int^1_0 \frac{dt~e^{xt}}{1+t^2} \sim \frac{\pi}{4} + \frac{x\ln(2)}{2} ~ \text{as} ~ x \to 0[/eqn]