Ostrogradsky method, integral of 1/(1+x^2)^2

[@renegado2630]

Omg when u made your cellphone say ostrogradski and then you said method i laughed so hard! Never heard of that method, mind blowing!

[@blackpenredpen]

santiago parra naranjo Hahahhahahahha. I like your comment!! Finally someone mentions about the iPhone. Hahahaha

[@TheGeneralThings]

Definitely my favourite method! Not as easy as trig sub, but wayyyy more mind-blowing.

[@mmunier947]

I just needed that for a Fourier transform proof and couldn't wrap my head around it Nice timing thanks!

[@neelu_16]

MMunier watch 3b1b for fourier intuition

[@mikenw255]

I just found this channel recently. You are very excellent and also entertaining. I wish I had you for a teacher when I was in college, I might have learned more. Thank you for your effort! Also ignore the detractors!

[@blackpenredpen]

Mike W I am really happy to receive your nice comment. Thank you, Mike!

[@holyshit922]

This method can be useful if 1) we dont have factorization of denominator (to find gcf we use successive divisions) 2) we dont know how to calculate trig integrals or forget some trig identities 3) we like partial fraction decomposition and we want to simplify integral before use this decomposition 4) we get integral from Euler or Weierstrass substitution

[@NoNameAtAll2]

can you tell more about successive devisions?

[@holyshit922]

In this example you have Q(x)and Q'(x) do the long division and remember the remainder R_{1}(x) and then take Q'(x) and R_{1}(x) to the next step (deg Q'(x)< deg Q(x) and that is why you choose Q'(x) to the next step) repeat this steps until remainder is some constant if it is zero last divisor is your gcf Maybe blackpenredpen will give you better explanation Do you know Euclid's algorithm with division for numbers ? This is similar

[@MujiburRahman-xk6vv]

The instructor is awesome. He should be commended for his excellent explanation of this beautiful method.

[@tarat.techhh]

it was actually cool the way you fast forwarded the video

[@martinzone8153]

Excellent. Just how the heck Ostrogradsky did come to that method?

[@peppybocan]

That's some Russian math school :D

[@holyshit922]

Key observation which allows you to derive this method is that rational part of integral appears if denominator has repeated roots (real or complex)

[@vyz333]

Amazing video! . Peyam's cameo at 2:53 made my day.

[@anatomania1126]

use the Chen Lu

[@shivanshnigam4015]

Nah, we gotta use lu Chen here

[@mohan153doshi]

Great way of doing this integral - and your teaching was uber cool - :)

[@blackpenredpen]

mohan153doshi thank you!!!!!!

[@adetokunboarogbonlo2491]

Great video blackpenredpenbluepen. Pls add the link to ``use the Chen Lu!``

[@blackpenredpen]

Adetokunbo Arogbonlo It's here on Dr. Peyam's channel! kzbin.info/www/bejne/b6PCZ52Cod2mmsU

[@holyshit922]

I think that another one example for this method will be useful , an example which shows advantages of this method when we have not given factorization of denominator explicitly

[@asifalamjoy9530]

You are a wizard . Thank a lot man

[@jameswilson8270]

I've never seen this. Awesome video.

[@blackpenredpen]

James Wilson thanks!!!

[@thephysicistcuber175]

Finally a method that I legit didn't know :)

[@holyshit922]

In my opinion this method can be simplified First calculate Q_{1}(x) = GCD(Q(x),Q'(x)) It can be done two ways 1) using factorization of Q(x) 2) using Euclidean algorithm for polynomials Then calculate Q_{2}(x) = Q(x)/Q_{1}(x) Calculate auxiliary polynomial H(x) = Q_{2}(x)Q_{1}'(x)/Q_{1}(x) Assume that degree of each numerator are strictly less than degree of corresponding denominator Solve system of linear equations written from equation below P(x) = P_{1}'(x)Q_{2}(x) - P_{1}(x)H(x) + P_{2}(x)Q_{1}(x) For an exercise prove that H(x) always be a polynomial

[@KeyMan137]

7:52 - 9:05 was my favorite part. Cat > Bunny > Oreo

[@ogusqiu6926]

so when do we use ostrogradski's method over trig sub and partial fractions?

[@thecorpooration]

This video is amazing! Thank you!

[@lkxemmeji7588]

Привет красный фломастер чёрный фломастер ! Классное видео с музыкальной паузой :)

[@dmitrii.zyrianov]

Oh, I hoped that it would be the method with some sort of using residuals and integral over closed curve

[@lionelinx7]

Damn bprp back at it with more supreme swag👌🔥🔥 and some fire integration techniques too

[@holyshit922]

People which like partial fraction and undetermined coefficients will like this method I know another way calculating this by using reduction formula Int(dx/(1+x^2)^n)=Int(((1+x^2)-x^2)/(1+x^2)^n dx)=Int((1+x^2)/(1+x^2)^ndx)-Int(x^2/(1+x^2)^ndx) Int(dx/(1+x^2)^n)=Int(1/(1+x^2)^(n-1)dx)-Int(x*x/(1+x^2)^{n}dx) First integral is reduced by canceling common factor , second integral can be reduced by calculating by parts u=x , dv=x/(1+x^2)^ndx du=dx, v=-1/(2(n-1))*1/(1+x^2)^(n-1) 1 can be replaced with a^2

[@divyanshukumar6807]

0:19 😂😂😂😂 it made me laugh so hard 😂😂😂😂 _😂_ _😂_ _😂_

[@dev02ify]

I did it with partial fractions. first do 1/(1-x^2)^2 (substitute ix later). since (naive partial fraction) 1/(1-x)^2 + 1/(1+x)^2 gives (2x^2 + 2) / (1-x^2)^2 the integral can be written integral[ 0.5/(1-x)^2 + 0.5/(1+x)^2 - (x^2)/(1-x^2)^2 ] First two terms are trivial to integrate. For the last term, we can integrate by parts: 0.5*integral[ x*(2x)/(1-x^2)^2 ] 0.5*( x * integral[ (2x)/(1-x^2)^2 ] - integral[ integral[ (2x)/(1-x^2)^2 ] ] ) For the integral[ (2x)/(1-x^2)^2 ] It's a chain rule, and gives 1/(1-x^2) So now, the integral of that last term is finally 0.5*( x/(1-x^2) - integral[ 1/(1-x^2) ] ) 0.5*( x/(1-x^2) - integral[ 0.5/(1-x) + 0.5/(1+x) ] ) 0.5*( x/(1-x^2) + 0.5*ln - 0.5*ln ) 0.5*x/(1-x^2) + 0.25*ln(1-x> - 0.25*ln Subtracting it from the integrals of the first two terms yields 0.5/(1-x) - 0.5/(1+x) - 0.5*x/(1-x^2) - 0.25*ln + 0.25*ln x/(1-x^2) - 0.5*x/(1-x^2) - 0.25*ln + 0.25*ln 0.5*x/(1-x^2) - 0.25*ln + 0.25*ln 1/2*x/(1-x^2) - 1/4*ln + 1/4*ln Now you substitute ix and divide the expression by i, and convert the complex logarithms to trig functions

[@holyshit922]

Yes but you used complex numbers

[@AndDiracisHisProphet]

That was almost not super much harder than trig sub^^ But nice, never heard of it. Also, cats>bunnies

[@blackpenredpen]

AndDiracisHisProphet how about cookies?

[@AndDiracisHisProphet]

It's an ordered set. cats>cookies>bunnies

[@blackpenredpen]

i must disagree.... cookies > anything

[@AndDiracisHisProphet]

well even you are allowed to be wrong, sometimes.

[@BulaienHate]

But I don't remember things, so every time I do trig sub I have to derive all of the trig Identities

[@holyshit922]

I have an example for next integral battle Int{\frac{dx}{x^2(4x^2-3)^2\sqrt{x^2-1}}} but with two different subsitiutions (fe inverse trig substitution and Euler substitution) I had got this integral from differential equation and this integral shows that inverse trig substitution is not always the fastest one

[@matthewzuelke6721]

Love the music!!!

[@MathIguess]

You assigned the value 0 to the variable C, then added C to the integral as the integration constant... :O

[@ms97]

Wow, amazing explanation. Thank you so much :D

[@jivjotsingh7579]

i wanted u to pronounce...!!anyways gr8 job my friend......Don t forget me...... this is harsimran singh here

[@jivjotsingh7579]

and i like the oreo with green halloweenyy eyes......oooooooooo!!

[@shakesmctremens178]

Sorry, dunce here.. don't recognize 8:32 - 8:52. What is the name of that method?

[@dlevi67]

It's just algebra. If you have 1 = Cx^3 + (D-A)x^2 + (C-2B)x + (A+D) for every value of x, then: C = 0 since there are no x^3 terms on the other side D-A = 0 since there are no x^2 terms on the other side => D = A C-2B = 0 since there are no x terms on the other side, and since C = 0 then B = 0 too this leaves us with A+D = 1, and since D=A from the x^2 term, both are equal to 1/2

[@shakesmctremens178]

So in other words, (?) because the x^3 coeff on the left is 0, then the x^3 coeff on the right, C, must also be 0?

[@dlevi67]

There is an equal sign between the two parts, so unless you can think of another, easier way of making them the same... };-) Sorry for being snarky; it really is simple, sometimes. Other times you need to solve a system of equations in the coefficients in a more formal way.

[@shakesmctremens178]

So is that a yes or a no? ..did I interpret you correctly or not?

[@dlevi67]

It's a yes. The two sides have to be made equal, and the only way to do so for all values of x is to set some coefficients to zero.

[@giampiproietti]

He uses the Hermite decomposistion for rational functions

[@holyshit922]

Russian textbooks claim that this method comes from Михаил Васильевич Остроградский but Hermite was only one generation younger and its possible that they found it independently

[@holyshit922]

There were many accidents like this for example in chemistry conservation of mass was first discovered by Михаил Васильевич Ломоносов and then rediscovered by Antoine de Lavoisier Sources claim that they discover it independently

[@emilioreyes7369]

@@holyshit922 Brother, I don't know if you will reply to this, but I am very interested in this Russian methods like Euler's sustitution, Weirstrass, Ostrogradski, do you know any book that has these methods detailed? Until now I've only found Piskunov's but it doesn't go into it very deeply

[@holyshit922]

@@emilioreyes7369 I have only Г.М. Фихтенгольц Курс дифференциального и интегрального исчисления because it is translated to Polish You can find other books if you abe able to speak Russian I was born too late to be able to speak this language

[@hoangdungvu3103]

thank you so much!

[@mohd.riyazkhan4515]

Can we take x=tan t and then proceed .We will get our final integral as integral of (cos t)^2 and then we can apply double angle formula (1+cos2t)/2 and solve the integral

[@Dhukino]

2:53 for epic peyam cameo appearence :D

[@nyehard3770]

Does anyone know the song that starts playing when he's is finding a common denominator around 8ish minutes?

[@johnhwhittaker6005]

2:53 literally shocked me to death

[@jamieee472]

Nice method!!

[@nathanaelmoses7977]

When partial fraction and integration by parts have a kid

[@arda3420]

As a high school student cant get anything stay strong trig sub!

[@tarat.techhh]

well i like these animal oreo more than the real one except when i am hungry ...

[@tarat.techhh]

ostrogradsky????? i don't even know what it means and ???? but still blackpenredpen is the best ...... thanks for uploading everyday

[@jozef_kascak]

Cheesysins It is a name of man who came with this method.

[@holyshit922]

if we solve this system of equation using matrix inversion we can change numerator and then calculate new integral easily

[@MrJasiekGuitar]

Do you know this integral: ∫√(a^2-x^2) dx from 0 to a? It's kind of tricky.

[@holyshit922]

Jasiek gdy chodziłem do szkoły to taką całkę liczyło się przez części (metoda całek stowarzyszonych) albo po prostu korzystając z interpretacji geometrycznej

[@lukandrate9866]

πa²/4?

[@animationtv1908]

Nvr heard of this method bt th moment i saw t,t automatically bcm my fav👍

[@MurakDurak]

only the real oreo man...

[@CornishMiner]

Great video. However, I do prefer it when you don't speed up the video. You've recorded it anyway, just play it normal speed please. Thanks :)

[@blackpenredpen]

Ok. I will keep that in mind.

[@TheGeneralThings]

I disagree. It made it sooooo much more entertaining (literally laughed out loud when the music started playing). Plus, as someone who already knows how to do partial fractions, it felt nice not having to go through all those steps when there are other videos for exactly that, but not many showcasing the (lets Google Translate say it) method.

[@budtastic1224]

0.5x speed should help

[@jameswilson8270]

0.05x times speed

[@jameswilson8270]

No but really, the speedups are good for me

[@cedricp.4941]

AMAZING 😃

[@AbhaySingh-wy6sj]

Hi From India

[@raulmendozagomez696]

Thanks!! saludos :)

[@shubhamgupta9601]

Honestly this method is so great, Really hope they teach this during JEE prep

[@flincocowalski149]

hello.good video. Could someone help me whit this ecuation? "x=sin(y)-y". i had pass almost 2 weeks trying to clear the "y" whit no succes. this ecuation come from a problem of electronic power circuit ,and, I have to calculate the Vrms for an arbitrary angle. but thas no much of a plroblem. later in the exercise they asking me to solve the equation for the oposit calculatrion which is find the" firing angle" for a given Vrms(root middle square Voltage).no load

[@pfeffer1729]

flinco cowalski I don't think that has an answer in terms of elementary functions.

[@flincocowalski149]

Ian .thanx for answer. But what it means? ,It can be solved in any way?

[@pfeffer1729]

It means I don't think the function y = f(x) can be written by adding, multiplying and composing sine, cosine, exponential, log, and rational functions.

[@flincocowalski149]

Oh.. so has no solution by conventional algebra. How about phasors or complex numbers?

[@pfeffer1729]

You might be able to write it in terms of the lambert W function... I suggest you just expand the Taylor series for sin x, subtract x and invert that.

[@lostwizard]

Obviously, Oreo is way better than Oreo, or even Oreo for that matter. Or maybe that was Zathras.

[@kingbeauregard]

I believe it's pronounced "Oscar Mayer".

[@kutuboxbayzan5967]

Well if x= tant makes very eaay

[@holyshit922]

He used substitution you mentioned in another video

[@liz_cherrele4302]

Hi pls help me solve this its urgent.. Indefinite intergral using ostrigradyski method (2+x) /(1+x²) ²

[@aryanks2167]

Damn I was learning this only recently and it's hard to find any integration technique you haven't done!

[@abd-elrahmanmohamed9839]

Nice 2:53 :D

[@blackpenredpen]

LOLL

[@abd-elrahmanmohamed9839]

can you explain this method , please www.nabla.hr/CL-IndefIntegralB4.htm#top

[@holyshit922]

@@abd-elrahmanmohamed9839 You cut the curve y^2=ax^2+bx+c with secant line Few sentences about derivation are in Fichtenholz book I created exercise to find suitable substitution using right triangles and angle bisector Lets draw right triangle and label its sides and angle as in inverse trig subs Draw bisector of angle complementary to theta - other acute angle Bisector will create another right triangle In this new triangle choose angle Pi/4+theta/2 and calculate its tangent

[@darcash1738]

Hmm. It seems like it will always be the power - 1, and then will put the integral portions with the remaining 1 power. [(a+bx^m)^n]’ = [n*(a+bx^m)^(n-1)]*mbx^(m-1) So the greatest common factor is simply the thing itself minus one power. Therefore, when dividing it, you will be left with 1 power in the denom, since its n - (n-1). Ultimately, whatever the factors are it will work like that where it is just the power - 1 on the outside, and then only a power of 1 inside the integral. If there are more things in the denom to begin with, eg x^5*(x^3+5)^6 it’s still P1/(x^4) + P2/(x^3+5)^5 + Integrals of (P3/x + P4/(x^3+5))

[@pengshi7183]

听到了背景音乐里的周杰伦和王力宏！

[@blackpenredpen]

Peng Shi 哈哈哈哈 是的！

[@pengshi7183]

blackpenredpen 臭味相投！

[@martinzone8153]

"Translate to english" does not help at all..LOL

[@pengshi7183]

Martin Zone lol we were talking about the bgm of the video, apparently we have the same taste in music 😂😂

[@kevinnncraftXD]

I did it integral by trigonometric substitution and i have other answer :c

[@fountainovaphilosopher8112]

Coooool

[@blackpenredpen]

Cooooool

[@darkahmedp]

yes

[@ApplepieFTW]

Pretty epic

[@aashsyed1277]

9:51 i was laughing

[@XDjUanZInHO]

Piskunov ;)

[@maurocruz1824]

Demidovich.

[@ishaanthakur7382]

i can pronounce ostrogradski!!!!!!!!!