evaluating the integral of 1/(x^4+1) from 1 to inf

blackpenredpen

[teavea]

This integral required a "green pen substitution" to be completed.

[moosemanuk]

The mind boggles. Thing is, the reason why I love your videos mate, is that I can follow each step easily (I have some university mathematics to rely on) ... but it's just the creativity of coming up with the next step that impresses me all the time! I would take one of those steps and while understanding it, would not even think of it in the first place. Really good stuff my friend, inspiring to an old bloke like me!

[not_vinkami]

14:03 this is the most important part of the whole video 😂 laugh die me

[vehaanhanda]

11:38 You could also have used the formula (integral)dx/(x^2-a^2) = (1/2a)ln|(x-a)/(x+a)| + C
That's a much quicker method, but the good part about your method is that....I finally got to know the derivatives for inverse hyperbolic tangent and cotangent :)

[Theraot]

14:00 I've been wondering how often that happens. Usually markers are alcohol based, and alcohol is volatile, and the two color switch technique requires to keep the markers uncapped. I read there are water based erasable markers, perhaps those last longer, I haven't tried.

[danieljose]

Awesome job man I really appreciate this channel and all the extra credit math I can do outside of school

[prollysine]

Dear bprp! I admire your soaring interest, I think you are standing alone in this round world! Thank you!

[gregoriousmaths]

That first step was genius! I was only able to do it after u did that

[mariokraus]

Physical pleasure in seeing such difficult integrals solved

[quantumcity]

Awesome video....😘👌 But I have another question about differential equation... That
🤔In 1715, 20 may, Leibniz revealed the solution of the differential equation x^2.y"=2y and I don't know how to find the solution of this differential equations can you please make videos on it ...🙋

[holyshit]

If we want to calculate numerically it is good idea to use change of variable u=1/x
then we will get Int(u^2/(1+u^4), u=0..1)
Answer is close to one quarter
If we need to have initial guess one quarter is good option

[seegeeaye]

it could be done in a short way. The integral = 1/2 integral of (1-x^2)/(x^4 +1) + (1+ x^2)/(x^4 + 1), the first integral could end up with a log function after letting t = 1/x + x, and the second integral could end up with an arctan function after letting t = 1/x – x, then consider the integral limits,

[whilewecan]

Thank you. I enjoyed your technique. With complex integral, it should be more mechanical to get the result.

[imanolmanzanares]

Nice video bro!! I am learning a lot with yours videos, thanks!

[rishavgupta]

Can you plz start weekly integral challenges

[i_am_anxious]

This is like a weirder best friend, who wants part of him to have a fourth power and part not, part negative of the original and part not, and his x wants to go from infinity to 1 instead of being trapped from -1 to 1. This best friend wants to explore his limits and see if he can go to infinity lmao

[admink]

Inverse ^-1
Hyperbolic h^-1
Cotangent coth^-1

[maxblanc]

Can we just put x² = u and just calculate the primitive of 1/1+u². Then we can write the primitive of this function as a limit [arctan(u) ] between 1 and infinite. Because the function is continue on I=[1, inf]
And the function 1/1+u², in +Inf, is equivalent to a Riemann function with t^-2 which converge
α=2 >1. So, by comparison theorem 1/1+u² CV. We can write the integral is equal to the Primitive's limit in +Inf minus the value of the primitive in 1. u = x² does not change the result: π/2 - π/4 = π/4

[_DD_]

Why not complexifying the integral? 1/[(1+(ix)^2)(1-(ix)^2)] then partial fractions and good game. 😊

[talharizvi]

U can replace integral of (1/u^2 - a^2) dx with 1/2a log | u -a / u+a| instead of using hyperbolic cotangent