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

[@teavea10]

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

[@rchishray7814]

teavea10 And a brown note

[@blackpenredpen]

YUP!

[@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!

[@tomatrix7525]

Stefan McNamara same thing here. I understand everything but always question how he thinks of doing in....

[@AlexisGarciator]

@@tomatrix7525 same here. I’m learning this stuff and understand it but sometimes question how to go about solving problems

[@antimatter2376]

Keepin' it ʳₑᵃₗ

[@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.

[@ianmoseley9910]

Alfonso J. Ramos Electronic whiteboard would do the job

[@not_vinkami]

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

[@blackpenredpen]

LOLLL Glad that you like it and I didn't cut that part off : )

[@p.singson3910]

Integration for breakfast is awesome

[@blue_blue-1]

Thought the same, but I don‘t make it to the happy end today. ☹️ Made it to the end now. Quite irrational, but fantastic handling. Kudos!

[@vehaanhanda1996]

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 :)

[@prollysine]

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

[@gregoriousmaths266]

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

[@mariokraus6965]

Physical pleasure in seeing such difficult integrals solved

[@danieljose4176]

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

[@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,

[@rishavgupta2117]

Can you plz start weekly integral challenges

[@holyshit922]

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

[@admink8662]

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

[@i_am_anxious02]

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

[@imanolmanzanares5000]

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

[@talharizvi6343]

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

[@Drk950]

It's the same :)

[@Gold161803]

The "so good" is back! YAY!!

[@KelfranGt]

How do I know when to use arctan(x) or arccot(x) for indefinite integrals?

[@blackpenredpen]

Kelfran Gt You can always use arctan(x) since its domain is all real numbers

[@KelfranGt]

blackpenredpen Oops! Sorry, I meant arctanh(x) or arccoth(x)

[@blackpenredpen]

@@KelfranGt Oh no worries. If -1

[@TheDanc000l]

@@blackpenredpen what if something like from 0 to 5 or something like that ?

[@WhattheHectogon]

@@TheDanc000l then right the integral as a sum of two integrals, with the bounds that you'd like

[@whilewecan]

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

[@joecamroberon9322]

You are a mathematics professor that wears supreme? You are a paragon of professors.

[@shlokjani9501]

Watching normally his videos(with blue and black pen) but you see blue pen entering(something serious is going to happen ) and at last green pen enters 🤯🤯🤯🤯🤯🤯🤯🤯🤯🤯🤯🤯🤯🤯

[@paulraj7573]

briliant

[@tomatrix7525]

I may be wrong, but did he make a small error at 13:30 . 1/1-x^2 is not actually (1/w^2+1)-1 . So he could nit change thr sign of the 1/2 outside the integral to positive.

[@quantumcity6679]

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 ...🙋

[@Demki]

here's a deceivingly simple integral, you could go down the rabbit hole of trig identities, or you could do it the "easy way": integral of (7sin(x)+9cos(x))/(4cos(x)+5sin(x)) dx The easy way would be to say "wouldn't it be nice" and express 7sin(x)+9cos(x) as A(4cos(x)+5sin(x))+B(4cos(x)-5sin(x)) for some constants A and B: 7sin(x)+9cos(x) = A(4cos(x)+5sin(x))+B(4cos(x)-5sin(x)) 7sin(x)+9cos(x) = (5A-5B)sin(x)+(4A+4B)cos(x) 7=5A-5B 9=4A+4B solving the linear system of equations with gives A=73/40 B=17/40 Then the integral becomes integral of [(73/40)(4cos(x)+5sin(x))/(4cos(x)+5sin(x))+(17/40)(4cos(x)-5sin(x))/(4cos(x)+5sin(x))] dx = integral of [(73/40) + (17/40)(4cos(x)-5sin(x))/(4cos(x)+5sin(x))] dx which can be solved by simple u-substitution.

[@blackpenredpen]

Demki : ) thank you. kzbin.info/www/bejne/fmOzpYJrlt57opo

[@Demki]

Oh I missed that video blackpenredpen :P I happened to come across a similar problem recently and thought of this channel.

[@blackpenredpen]

Demki ah I see. That's a really fun integral.

[@deathmetal4029]

To them who find this problem difficult , you might get disappointed but this problem is fuckin easy, its just 4-5 min problem. Well that my opinion. -Jee aspirant,Indian.

[@raytheboss4650]

You can take lcm in denominator

[@goodplacetostart9099]

So I learnt that (a-b)^2=(b-a)^2 When I tried the problem on my own😂

[@patrickgleason3053]

You just blew my mind 🤯

[@blackpenredpen]

: )))))

[@osuNoobCast]

14:01 God doesn't want it to be a plus 😜

[@ackheeleangus1592]

The math was great but is no one gonna give him props for the supreme windbreaker 😂

[@flamingpaper7751]

What about the same function from negative infinity to positive infinity

[@benjaminbrady2385]

How do we know what infinity in the trigonometric functions are?

[@pent_m8589]

I think you can also use the secant substitution

[@punkrockrules205]

great techinique

[@Jacob-uy8ox]

Best video of 2018

[@blackpenredpen]

Thanks!

[@abstruse123]

Excellent explanation

[@maxblanc1934]

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

[@lawliet2263]

Oh my god what a pro dude

[@dillonlobban1922]

Beautifully done

[@dannyvillegas9104]

You can also use integration by parts

[@omar.ma7]

An error in 6th line :in denominator we have: u^2+2=2(u^2/2 +1)=2((u/sqrt(2))^2+1) Not sqrt2(u^2/sqrt2 +1)

[@Drk950]

Question: Would be Feynman technique useful in this problem?

[@kizki2351]

Hey can you please link me to one of your videos explaining the 11:02 step / or someone can please explain to me ? I don't understand this step with tan-1. Thank you

[@efegokselkisioglu8218]

Yeah, me neither

[@_DD_15]

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

[@JoshuaHillerup]

Are there integrals that you can do with us that just won't work in Wolfram Alpha?

[@M.Neukamm]

That's the funkiest integration exercise I ever saw. My mind is boiling.

[@pdjibril]

😂

[@manishkumarsingh3082]

Solve it x+√x(x+1). +√x(x+2). +√(x+1)(x+2). =2 for positive value of x

[@megauser8512]

Assuming that each of the 2nd through 4th terms are entirely under their square roots, the solution is as follows: Given x + √x(x+1) + √x(x+2) + √(x+1)(x+2) = 2, x > 0 , then x + √(x^2+x) + √(x^2+2x) + √(x^2+3x+2) = 2, x > 0. Plugging in x = 0 we get: 0 + √(0^2+0) + √(0^2+2*0) + √(0^2+3*0+2) = √(0+0) + √(0+0) + √(0+0+2) = √0 + √0 + √2 = 0 + 0 + √2 = √2, which is < 2. Plugging in x = 1 we get: 1 + √(1^2+1) + √(1^2+2*1) + √(1^2+3*1+2) = 1 + √(1+1) + √(1+2) + √(1+3+2) = 1 + √2 + √3 + √6, which is > 2. Notice that all 4 terms always increase with x when x is > 0. Therefore, the solution must be x = some number between 0 and 1. Going further, x + √x(x+1) + √x(x+2) + √(x+1)(x+2) = 2 turns into √x * [√x + √(x+1)] + [√x + √(x+1)] * √(x+2) = 2 [√x + √(x+1)] * [√x + √(x+2)] = 2, and that's all I got. However, Wolfram Alpha says that the answer is 1/24. (1/√24 + 5/√24) * (1/√24 + 7/√24) = (6/√24) * (8/√24) = 48/24 = 2.

[@dark_knight2341]

Hey , why don't you solve this integral using a trig sub : we put x = sqrt( tan(u) )

[@absolutezero9874]

Hmm just leave the inverse cosh of square root of 2?

[@absolutezero9874]

It’s equal to ln(square root of 2 + 1)

[@VSP4591]

Why not to split x4+1 = (x2+i)(x2-i) and split in 2 simple fractions and continue to integrate 2 fractions of second degree 1/(x2+i) and 1/(x2-i).

[@__________________________6744]

do u have all supreme clothes?

[@johnpaulbdeluna]

can I answer this using arctan?

[@ianmoseley9910]

Any mileage in x^2+1 times x^2-1 = x^4-1?

[@djaziameziane8386]

I think that the first integral in blue is 1/2 arctan(u/√2) and not 1/√2arctan(u/√2)

[@ruathak1106]

14:06 - "You seein' this shit?"

[@nathanisbored]

if arctanh(x) and arccoth(x) have the same derivative, does that mean they are only off by a constant? is that an identity?

[@blackpenredpen]

I actually have a follow up video already, see description.

[@MarioFanGamer659]

The graphs of arctanh(x) and arccoth(x) don't really look similar, although they start and end at the same points so I guess you can say they togther extend the domain of the integral of 1/(1-x²) with x = ±1 being a singularity. That being said, their sister functions arctan(x) and -arccot(x) (notice the minus) *are* off by a constant (although the constants depends on the value of x because arctan(x) + arccot(x) is π/2 if x ≥ 0 and -π/2 if x < 0).

[@alextaunton3099]

@@MarioFanGamer659 is the singularity at just the value where re=1 or -1 and the imaginary value is 0, or for all imaginary values for real value 1 or -1?

[@MarioFanGamer659]

@@alextaunton3099 As a rule, if something isn't mention, it's implied to be zero, not any value. In fact, this is quite consistent, I don't know how you came to the idea I implied _all_ imaginary numbers (i.e. a line of singularities) and not just two values with no imaginary component (i.e. two points).

[@alextaunton3099]

@@MarioFanGamer659 because you used the term "singularity" which usually is used in context of complex functions

[@dmytro_shum]

You can also do that: coth^-1(2^0.5) = ln (2^0.5 + 1) I have made that in 1 minute

[@dmytro_shum]

The most fun thing is that wolframalpha gives it in the form of: 1/2 log(1 + 1/sqrt(2)) - 1/2 log(1 - 1/sqrt(2)) =D

[@zanna5333]

How about sqrt of (e·sqrt of(pi · sqrt (e·sqrt (pi· sqrt ....()))?

[@mike4ty4]

How about now int 1/sqrt(x^4 + 1) dx from 1 to oo ?

[@blackpenredpen]

We will get a sad face : (

[@mike4ty4]

@@blackpenredpen Why?

[@blackpenredpen]

mike4ty4 We can only do approximation in that case. I don't think we can find a closed form for its anti derivative

[@mike4ty4]

​@@blackpenredpen 1. This is a definite integral, not an antiderivative, using the same bounds as used in the video. 2. Wolfram says the value of the integral is 2 Gamma(5/4)^2/sqrt(pi) ~ 0.92704, using the gamma function, which has featured on this channel before. It would be interesting to see how that is actually obtained by hand.

[@jeremyb1346]

[EDIT: I thought the integral was on R⁺ LOL ,anyway, using u-sub u=1/x, we can show that the integral on R⁺ is 2 times the integral on [1, infinity) @@blackpenredpen Hi (from france btw xD) blackpenredpen and mike4ty4, for the defined integral, here’s a BEGINNING of proof, using Ramanujan’s Master theorem: en.wikipedia.org/wiki/Ramanujan%27s_master_theorem Let’s call F(x)=sqrt(1/(1+x⁴)) First notice that the mellin transform of F is nice : www.wolframalpha.com/input/?i=mellin+transform+of+1%2Fsqrt(1%2Bx%5E4) We can define (R) as : (R) : mellin transform of F = gamma(s/4)*[gamma(1/2-s/4)/(4*sqrt(pi))] and M(s)=mellin transform of F Considering s’=s/4, using u-sub u=t⁴ such that we can have a term in u^(s-1) we have : M(s’)=1/4*mellin transform of 1/sqrt(1+u) = M(s) www.wolframalpha.com/input/?i=mellin+transform+of+1%2F4*sqrt(1%2F(1%2Bx)) We want to prove the relation : (R’) : M(s’)= gamma(s’)*[gamma(1/2-s’)/(4*sqrt(pi))] And then we want to compute the value : M(s’=1/4) which will give us the answer. (R)(R’) for some values of s (convergence for the values we are interested in, can be shown with Riemann, for x==> infinity ; 1/x^alpha converges alpha >1) It turns out : sum of (-x)^k / k !*[gamma(1/2+k)/(4*sqrt(pi))] from 0 to inf= 1/sqrt(1+x) for |x|

[@s.p.a.3583]

I'd surrender: can u invert f(x)=2+ln(x)+x? Nothing program like geogebra can't too.

[@holyshit922]

If we prefer artanh we can use artanh(1/x)

[@blackpenredpen]

then.. i need another video to follow that up : )

[@robydomp]

Is w = X+1/x so for X =2 w = 5/4

[@vishalmishra3046]

You did not finish to the end of simplification of final answer. arc cot h (x) = 1/2 ln ((x+1)/(x-1)). So, arc cot h (sqrt(2)) = 1/2 ln [(sqrt(2)+1) / (sqrt(2)-1)] = ln(sqrt(2)+1). So, final answer = [ pi/sqrt(32) + ln(sqrt(2)-1) / sqrt(8) ] = same as your answer but much simpler without "inverse cot h". Use cosh(x) + sinh(x) = exp(x) to compute inverse-cot-h.

[@hjk0150]

I came up with π√(2)/8 - √(2)/4*ln(√(2)+1) and it seems to be the same. Not sure since I haven't learned about hyperbolic functions yet

[@matthiasheymann]

One can compute z:=coth^{-1}(√2) by writing √2=coth(z)=(e^{2z}+1)/(e^{2z}-1), multiplying both sides by e^{2z}-1, solving for e^{2z}, and then taking the log to solve for z. He should have done that instead of leaving his final result as he has, with coth{-1} still flying around.

[@EduardoHerrera-fr6bd]

The integral from 1 to inf of 1/(x^4+1) dx - Hardcore Partial Fractions

[@EduardoHerrera-fr6bd]

Oh, this was for the other video :(

[@haidarkhalid7238]

dos any body know what the name of this person i want to contact with him

[@l.g.f.c7356]

You can help me with this equation : x^(1/x) = i I don't know how yo use the Lambert W function un this ecuation

[@blue_blue-1]

L.G.F.C Equation

[@l.g.f.c7356]

Ok :v

[@hypnovia]

Take 1/ both sides

[@KnakuanaRka]

X^(1/x)=i 1/x^(1/x)=1/i=-i 1/x*ln(1/x)=ln(-i) ln(1/x)*e^ln(1/x)=ln(-i) ln(1/x)=W(ln(-i)) 1/x=e^W(ln(-i)) X=e^(-W(ln(-i))) Good luck evaluating that.

[@jakemoll]

L.G.F.C Raise both sides to the power of x: x = i^x Then take ln of both sides: ln(x) = x ln(i) x = e^ln(x) so we can write ln(x) = e^ln(x) * ln(i) Divide both sides by -e^ln(x): -ln(x) * e^-ln(x) = -ln(i) Now we can use the W function because we have something of the form z * e^z: -ln(x) = W(-ln(i)) so x = e^-W(-ln(i)) ln(i) = i*pi/2 so x = e^-W(-i * pi/2) Hope this helped

[@warrickdawes7900]

0.33775542942802... I'm glad Wolfram Alpha knows!

[@anupamkhatiwada7719]

if i am allowed to plug in any formulas for complex numbers then i got the answer to be pi/4(3+i)

[@arturrheinboldt2207]

That shit the realest yo

[@hunterhunter4099]

thanx bro

[@krishbishwanath]

Easy question 😁😁

[@daniell321]

can you cover regular and singular non linear ODE pertubation theory?

[@rosarioromano8550]

stai semp buon

[@rohithc681]

Wow aweaome

[@lucasargandona4658]

pfft I learnd that in elemnatry skool./

[@alexanderskladovski]

just approximate x^4+1 as x^4 at big x...

[@harishpanwar6737]

Hello sir Im harish panwar from delhi Plz can you solve An integral I= 1/1+sin^4x

[@miguelhermida9072]

Uf men, that integral

[@takyc7883]

AYO BPRP GOT DRIP

[@rishichhetri3112]

😒😒Yes Pretty fun

[@jaredbeaufait5954]

I feel afraid

[@Pineapplecake]

Nice shirt!

[@blackpenredpen]

Thank you!!

[@danielmacsai776]

solve sqrt(x+5) = x^2-5 good luck:)

[@i_am_anxious02]

Sqrt(x+5)^2=(x^2-5)^2-> x+5=x^4-10x^2+25-> x^4-10x^2-x+20=0 So x is either -(1/2)-sqrt(17)/2, (sqrt(17)-1)/2, 1/2+sqrt(21)/2, or 1/2-sqrt(21)/2

[@danielmacsai776]

u sure about that?

[@i_am_anxious02]

Dániel Mácsai Yeah, just square the dare root. I mean, if you want only the square root being positive, fine, but then you just like to nitpick. Squaring cancels with the square root, and I used the polynomial of (a+b)^2=a^2+2ab+b^2 to solve (x^2-5)^2.

[@danielmacsai776]

Yeah, only two of those are right The challenge here is to find the answers without computer/quartic formula. Try it with only paper and pencil:)

[@i_am_anxious02]

Dániel Mácsai what quadratic formula? It’s a quartic equation.

[@VanNguyen-kx6gx]

1 another resolution.

[@eboone]

hello

[@neilgerace355]

More backwards writing :)

[@x-warrior8517]

i solve it by using trigonometry.Thanks ❤

[@Drk950]

How?

[@jblac201]

🖋🖍 Do ∫(3x³-x²+2x-4)/(√(x²-3x+2))dx from 0 to 1 🤣🤗🤔

[@lamnguyenthanh6178]

De hiểu rõ

[@zuccx99]

🗿

[@Hossein518]

i have a better way to solve this .

[@calvinjackson8110]

Very tricky and not direct. Very non intuitive. The majority of us do not think like this. Most of us would not see to do this slight of hand approach. Unfair problem. Only someone like you could be expected to see to do this.