life changing integration by parts trick

Let’s learn a life-changing integration by parts trick. Once you learn this integration technique for you calculus 2 class, many integrals will be much easier. The trick here is to choose a clever antiderivative with a smart constant.
0:00 Intro
0:15 Integral x arctan x
2:20 Integral ln x+2
3:30 Integral arctan square root x+1
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[@ThJakester897]

This technique is absolutely crazy. I feel like I now have the confidence to ask my boss for a raise and find a beautiful wife to marry. Thanks, Dr. Peyam.

[@idk-dr7fo]

lmao

[@stevenwilson5556]

but, be careful, if you watch the video backwards, it turns into a Country Western music video…

[@oswaldozanella3541]

LMAOOO

[@watching-machine69]

LMAO

[@YU-vz7bf]

LMAOOOOOO

[@blackpenredpen]

This is actually really cool 😆 Glad I actually watched it haha

[@drpeyam]

Awwwww thank youuuu!!!

[@borg972]

Please combine this with the DI method!😺

[@ILoveMaths07]

Nuuuuu! Now bprp will kill his students with these questions!

[@petereziagor4604]

Master 😅

[@pain8117]

What is this crossover episode? 🧐

[@DrTrefor]

How did I not know this. My life is changed.

[@tanish6035]

Greetings sir

[@kiit8337]

Orz here

[@ayan7bhowmik]

It's a very normal question in WBCHSE class 12

[@divinecreation7429]

​@@ayan7bhowmikNo one teaches this method in class 12

[@nonentity168]

Now the +C is actually useful. Great video

[@whannabi]

Yup

[@jaskhetani4631]

So trueee

[@liamloveslunch]

"Arent you glad I found my video? 😁" Yes! So happy! Glad to see you so happy about math too :)

[@gamingroom2541]

3:11 “Isnt that nice!” Nice ? NICEEE ??? SIR THAT IS STRAIGHT AMAZING my life is never the same after this

[@sharpnova2]

I've taught this method to my students for years and it's always made them much better at integrating in general

[@aarjith2580]

Hey, I have a doubt, Why do we have to choose the same constant at both the places though? for example at 4:47

[@agustincai]

symbolab gave me another result in his first integral

[@kilian8250]

A Arjith look at the proof of integration by parts

[@maximlavrenko1164]

@@aarjith2580 because we can't choose different ones. We are plugging in the antiderivative in both places you bum

[@xinpingdonohoe3978]

​@@aarjith2580 you integrate 1 and differentiate arctan(√(x+1)). Your derivative of that term is set, but the integral of 1 is x+c. This works for any fixed c. Hence, we fix c=2 as it's most useful to us. This means, each time we use this specific antiderivative of 1, our constant is 2.

[@krischalkhanal2842]

I remember asking teacher, "can we add a constant to *inner integration* in the formula of integration by parts?" He told me, "only a single +c can appear in a single integration", which feels so stupid to me now. I am so glad I found this video. Kudos to you Dr. Peyam.

[@skylardeslypere9909]

I think he probably thought you meant a general +C when applying the formula, which I'd also perhaps not recommend because it'll look too messy. But specific constants to make it easier, definitely do it!

[@sharpnova2]

I'm sure it was more a result of you asking your question unclearly

[@MrSamshy]

Probably your question wasn’t clear to him, so he meant you don’t add +C within each term.

[@jakolu]

@@skylardeslypere9909 that's still a bad answer from the teacher imo, especially without qualification. Of course many teachers won't get into the specifics of derivations and logic because of their effort,. communication skills, or rigid worldview

[@mesaplayer9636]

I think he meant you were asking if you can leave constants in the equation without absorbing them into +C

[@floydmaseda]

Thought the title was click bait. It was NOT click bait. Life changing, indeed!

[@drpeyam]

Omg hi Floyd!!!

[@aarjith2580]

@@drpeyam Hey, I have a doubt, Why do we have to choose the same constant at both the places though? for example at 4:47

[@plislegalineu3005]

@@aarjith2580 When you're doing integration by parts, you integrate "dv", and you can add any constant to the v. But if the constants are different we would get v≠v

[@sayantansaharoy6349]

I love his attitude, man knows math and is happy about teaching it. I wish all my teachers were like him.

[@aaronpahler5410]

As a physic student I can say, this is really lifechnaging, hours of pain will be joy instead. Thanks 🙏🏼

[@emoney1uz]

blessed to have him as my differential equations professor, couldn’t be luckier

[@drpeyam]

Awwwww thank you ❤️ It’s symmetric!

[@kobethebeefinmathworld953]

The fact is (even if we just think up to calculus level): whenever we do antiderivates, 99% of the time we forget the "+C", and this "+C" is the reason why we can end up with different answers depending on how we choose. Moreover, this also shows that the operation of antiderivate is not even a function as we can pick anything in the codomain for C as we want. So as long as the "number" that we "need" belongs to the codomain we can add it freely.

[@gabrielmello3293]

Well, it is a function if you consider an equivalence relation between functions that differ by a constant.

[@gabrielmello3293]

(And then it is a linear isomorphism)

[@Cyrusislikeawsome]

👆 put differently, Give it a couple years and you'll stop thinking of constants as really making them "different"

[@MrTomas7777]

Wow, such a simple idea that's hidden in plain sight! Just in time for my calculus test this week.

[@Cjendjsidj]

Good luck for your exam bro.

[@skylardeslypere9909]

To be fair, the integral of x²/(1+x²) is relatively easy because that's just (1+x²)/(1+x²) - 1/(x²+1) which gives x - arctanx But nonetheless, this is a really useful trick I've never thought about. Nice video

[@sergioconfero6049]

Yeah I ve thought the same thing

[@tonyhaddad1394]

Me 2

[@user-th3tg4fb5g]

yeah i agree but this does seem to be a nice tip

[@annaclarafenyo8185]

Nope. The 'trick' only works in cases where the algebraic simplification also works. This is not a method, it is nothing.

[@Cyrusislikeawsome]

@@annaclarafenyo8185 the situations where OP's method work are precisely the situations in which Peyam's work. They're the same thing, just hidden behind algebra, like when an FB videos says to "take a number, double it, add one, half it, take away half the original number and boom you get 1/2 every time, spooky, right?"

[@MuPrimeMath]

How have I never seen this before? This is amazing!

[@drpeyam]

Omg hi Mu Prime Math !!!

[@imperiumgraecum9126]

this is some serious math hack territory we are treading in

[@violintegral]

This is one of the reasons why integration by parts is my favorite (and in my opinion, the most powerful) integration technique. Here's a really great (and difficult) integral for anyone interested: ∫√(1+x+x²)/(1+x)dx. Keep IBP in mind!

[@davida2810]

int (1+x+x^2)/(1+x) dx = int 1 + x^2/(1+x) dx, then use u = 1+x.

[@violintegral]

@@davida2810 no, that's the wrong function. (1+x+x²) should be under a square root. So the function was √(1+x+x²)/(1+x), which is much more difficult to integrate. Of course I wouldn't ask such a simple question that doesn't even involve IBP on a video about IBP.

[@Vladimir_Pavlov]

It is unlikely that integration in parts will lead to a result in this case. t=x+1 . ∫√(1+x+x^2)*dx/(1+x) =∫√(t^2 -t+1)dt/t=∫(t^2 -t+1)*dt/t*√(t^2 - t+1)= =(1/2)∫(2t-1)dt/√(t^2 -t+1)-(1/2)∫dt/√(t^2 -t+1)+∫dt/t*√(t^2 -t+1)=(*) In the last integral, replacing t=1/z leads it to an integral similar to the second. (*)=∫d(√(t^2 -t+1)) - (1/2)∫dt/√[(t-1/2)^2 +3/4] - ∫dz/√[(z-1/2)^2 +3/4]= =√(t^2 - t+1) - (1/2)* ln[t-1/2 +√t^2- t+1]- ln[z-1/2+√z^2-z+1]+C= = √(1+x+x^2) - (1/2)*ln[x+1/2 +√(1+x+x^2)]-ln[1/t-1/2+√(1/t)^2-(1/t)+1]+C= = √(1+x+x^2) - (1/2)*ln[x+1/2 +√(1+x+x^2)] - ln[(1-x)/2 +√(1+x+x^2)]+ ln|x+1|+C. (I use the equality ∫dt/√(t^2+λ)= ln|t+√(t^2+λ) |+C, a "long logarithm", and not a little informative expression through asinh).

[@violintegral]

@@Vladimir_Pavlov actually, after substituting 1/y=x+1, you can perform integration by parts with u=√(y²-y+1), dv=-1/y²dy. You actually did the same substitution (in two steps), but used a different method after that. Anyway, it's nice to see an alternate solution to mine. Reciprocal substitutions can be very powerful, along with ibp!

[@48_subhambanerjee22]

@@violintegral noicely done bro ... Yep we can use substitution.. but don't forget dy 😉

[@stevehof]

This blew my mind! So fascinating. I guess that's why in my courses it usually says "find the most 'general' antiderivative."

[@vrtxactivewear7383]

I've learnt the +1 -1 trick. on integration but this takes it to a new level - can't thank you enough!

[@gillesphilippedeboissay109]

Me too

[@sooryanarayana3929]

This trick is amazingly amazing, this really changed my life and how I see Integrals, thank you Dr. Peyam!

[@SelvaKumar-th1lh]

I don't even know integration 😂

[@sebastiangudino9377]

There are two basic techniques which are pretty easy to understand in isolation. One is integration by parts (Multiple interations of it are done with the tabular method, or DI method) the other is U-Substitution, which is like anti-chain rule. You learn those two bad boys, and if you are good with algebra and can also think geometrically, then you have a solid foundation to learn calc I-II ideas

[@jnfkjbfizbki]

I gotta say I love learning math from people who talk about its tricks and properties with such passion. Thank you for your video!!

[@Unchained_Alice]

I never even considered this. Makes it so much easier!

[@sankalpranjan1863]

Wow. As a math lover, I've fallen in love with this constant addition trick. As if, it was always there but we didn't see it ! Thanks a lot ❤️❤️❤️❤️

[@ldc0322]

I’m in my first year in college studying physics AND OH MY GOD I’M SO HAPPY I FOUND THIS BEFORE MY ANALYSIS EXAM

[@INDABRIT]

This is mind-blowing I love it, never thought of taking advantage of the constant like this. Nice!!

[@slavinojunepri7648]

Beautiful tricks never cross one's mind until someone with enough insights reveal them. Thank you Dr. Peyam!

[@uncertainukelele]

Me: has used integration by parts for years. Also me: never thought to use an antiderivative other than the C=0 version. Thank you for teaching me 😁

[@tomctutor]

Never thought of doing that, gonna call it the "Peyam Integration Method" Something quite related of course is shifting the function, so; ∫ f(x+c)dx = ∫ f(u)du letting u= x+c. Ex: consider ∫ x/(x+1)dx ➙ ∫ (u-1)/u du with u= x+1, gives ∫ (1-1/u)du = u- lnu = (x+1)-ln(x+1)+const. Be careful though to change limits in definite form, so ∫ f(x+c)dx between x=a to b ➙ ∫ f(u)du between a+c to b+c

[@drpeyam]

Yep this would also make the ln(x+2) integral easier too

[@tomctutor]

@@robertveith6383 The area under 1.x yes, but simply ln(x) for integral, thats what Wolfram Alpha says anyhow.

[@timothyaugustine7093]

It's inspired from Ross's analysis book.

[@coerciasink]

isn't that just substitution?

[@tomctutor]

@@coerciasink Indeed but works in general for any function f(x+c) -> f(u)

[@user-qd4kt7ze3o]

My lord, such a simple yet elegant trick and it's so convenient. Many thanks for sharing it with us!

[@funkalunatic]

My life has truly been changed. Thank you , Dr Peyam!

[@ashokvajpeyee8947]

Just can't explain how grateful I am to you for explaining this miraculously magical method Thanks alot ❤️

[@egillandersson1780]

What an amazing trick !!! Thank you for sharing.

[@antoniomacaione203]

It s amazing how something so important and so simple ,isn't used often

[@saulivor2843]

He knows how amazing this trick is and he is absolutely right, very glad I watched, thank you so much for the integration skills

[@cody5535]

Absolutely stellar expedient method!

[@8896tim]

For the ln(x+2)dx integral, you can substitute u=x+2 as well to bring it to a standard form

[@Panda_2397]

He is using the same method but not showing you the substitution. d(x+C) where C is a constant

[@carultch]

Here's an example with ln(), that is more interesting. integral ln(|x^2 - 4|) dx

[@zombiefelice]

Im almost crying how good this is , Thank you very much for sharing

[@yoroida]

You're the man!! Usually it takes more than one videos to get me to subscribe.

[@mudkip_btw]

Very nice trick! It's always easy to forget the simple things if you're doing hard problems. This might save me half an hour sometime in the future :D

[@nedmerrill5705]

Truly exciting and imaginative!

[@chrstfer2452]

Yeah that is pretty awesome. I definitely remember being taught this now that you mention it, but i didnt come up with it when i did the example so thanks for the reinforcement!

[@GrandAdmiralMitthrawnuruodo]

Wow. This just blew my mind! That‘s crazy cool!

[@kai_enforcer]

I don't know why, but this makes me really happy 🤤🤗🖤🖤

[@mathbattles1471]

Me 2😚

[@silentintegrals9104]

Solving integrals is allways fun!!!

[@mohammedal-haddad2652]

Dr. Peyam: Thanks for watching. Me: Thanks for changing my life!

[@zereentasnimmridula6006]

This is dope Please keep making integration easy and fun!! I need it badly

[@hornet1904]

Truly life-changing. Was absolutely mind-blown

[@Handelsbilanzdefizit]

Very good, and very informative. Integral of ln(x+c) is easy, by integrating the inverse y = e^x - c, and subtract from the total rectangle area (x+c)ln(x+c). So, in your case: (x+2)ln(x+2) - x + c But don't forget to substitude the intervall endpoints. That's how I did it for years ^^

[@tomctutor]

That's Laisant' method, quite smart really. Problem is you may not have the actual inverse function to work with (logs are easy of course) but what about 1/(x^3+x+1)?

[@michaelgolub2019]

Have never seen the method before... Really cool! I also remember than some functions (Ln, Arctan) are the same in the complex domain. It would be interesting to see similar simplification methods based on moving to the complex domain.

[@carultch]

Another method is to simply add zero in a fancy way, to reconcile the fraction within the integral. Starting with: x^2/2*arctan(x) - 1/2*integral x^2/(x^2 + 1) dx Add zero in a fancy way, to the numerator of the integral, by both adding 1 and subtracting 1: integral x^2/(x^2 + 1) dx = integral (x^2 + 1 - 1)/(x^2 + 1) dx Regroup: integral [(x^2+1)/(x^2 + 1) - 1/(x^2 + 1)] dx integral [ 1 - 1/(x^2 + 1)] dx = x - arctan(x) Combine with the original expression: x^2/2 * arctan(x) - 1/2*(x - arctan(x)) And simplify: (x^2 + 1)/2 * arctan(x) - 1/2*x

[@vazn4143]

This is amazing ! Definetly life changing in some ways, especially since i have my integration final soon :D Also i just saw i wasn't subscribed yet but im litterally binge watching all of your content, so i clicked the button this time ! That's somme great content !

[@brainpowerrb3003]

Pretty cool method. Never thought of it that way. This is pretty helpful. Thanks so much for sharing this

[@yogeshwagh2849]

Life changing man, for real 💯

[@silentintegrals9104]

Nice!!!

[@MarcusCactus]

At 0:50 he says "you start to cry necause this is very hard!" Well, not really... x² ÷ (x² + 1) is easily written as 1 - 1 ÷ (x² + 1) which integrates immediately to x - arctan x.

[@lukebennett9986]

Sorry legend

[@XTREMEShaurya-mg8xq]

Yea its easy

[@winter_c]

This helps a lot with both definite and indefinite integral. It saves lots of calculating time.

[@FlummoxTheMagnificent]

That was actually incredible.

[@syedzainulabidin5498]

Mann that boom boom 3:01 !!! was so energetic 😅🎉

[@silentintegrals9104]

Solving integrals is allways fun!!!

[@kayleeb6197]

I’m just now learning integrals and I’m so happy I found this

[@longjump123]

Keep up the good work. I love it. That trick is awsome!

[@jose6183]

I can't believe my eyes. Why didnt we know this before?

[@surajjani4868]

This is so awesome!

[@silentintegrals9104]

totally agree

[@EricaGunn-on7nf]

This is amazing, I always struggled with this but this helped so much!

[@chromiyum6849]

Totally blown my mind! Shared to all my calc homies

[@stevewhisnant]

Does this work with the D-I method of integration where you might have several levels of differentiation and integration? Here, if we are permitted to add an integration constant at each stage, we can build up a polynomial if required. That would be really cool and could really simplify some nasty integrals.

[@user-ls8fw6zm3j]

Why not? Any antiderivative is equally legit, isn't it?

[@drpeyam]

Of course! When you choose the antiderivative in DI, just pick the right constant

[@anupghimire4627]

@@drpeyam yah

[@carultch]

It can work, but it generally helps you the most on the regrouper types of integrals that you usually can do in just one row.

[@TheZanzaroni]

From my (limited) experience in math, if you just add and subtract that constant divided by the denominator in each case you should end up with the same result, still pretty fun though

[@timothyaugustine7093]

Yes, by adding a zero, you can obtain the same result as the trick although you might need to do a couple more steps.

[@zeddyknight2478]

I found this when I’ve been suffering the most with Calc 2 thank you kind sir

[@changyauchen]

This is helpful! Somebody might say that you still get the answer by the old way, but this new trick saves one step at least.

[@_wetmath_]

this trick is pretty cool, but after trying this in all my integration by parts questions, i can tell you that this trick only works for certain integrals and it's not something you should look out for all the time. just keep this trick in mind if you can cancel your antiderivative with a denominator later on

[@uwuifyingransomware]

How do you figure out which integrals it does/doesn’t work for?

[@Cjendjsidj]

@@uwuifyingransomware im guessing by practice

[@_wetmath_]

@@Cjendjsidj yeah by practice. it's not worth it to intentionally leave a gap and write (x+_). i suggest using this trick only when you are stuck or happen to notice the perfect setup for this trick

[@timothyaugustine7093]

@@uwuifyingransomware You'll figure it out by experience after dealing with a bunch of different functions in many integration problems. You must have substantial experience in functional analysis and how to manipulate those functions using several derived formula which will help simplify the problem. You may not be able to solve all integrals since not all integrals can be solved but most can be solved using the normal techniques you've learned i.e. elementary functions, algebraic/trigonometric/hyperbolic substitutions, powers of trigonometric/hyperbolic functions, recurrence relations, partial fraction decomposition for rational functions and of course, integration by parts.

[@LotkroMath-zy4me]

This feels illegal

[@julianmoody5939]

This video was recommended to me at midnight the day of my advanced integration unit test which is basically just a bunch of integration by parts and partial fractions. I’m both grateful and very concerned that youtube recommended me this

[@boukraailyesali9131]

my teacher taught me this trick before and I was like wow, thank you. Because of your video now I remember a useful trick to apply in my life

[@pandugus1550]

5:13 My favorite part of this video 😎

[@gunhasirac]

TBH without adding the constant term in the first place, I will still tackle the second term by adding 1 to the numerator which gives the same result with one more line, but this method is definitely more efficient and some thing I should keep in mind. Thx Btw second one I will do with change of variable, same as third, but third one still need a constant to do neatly, so very clever examples for showing the power it does.

[@user-er2nx5py7r]

Totally life changing trick , Thank you!!

[@nightshark1156]

"Speaking of Ellen DeGeneres...", I lost it, this may be the funniest math video I've seen. And the mind-blowing trick was icing on the cake. Job very well done Dr. Peyam, thank you.

[@rarebeeph1783]

so basically, this is taking the "integral(udv) = uv - integral(vdu)" and expanding it without loss of generality to "integral(udv) = u(v+C) - integral((v+C)du)"

[@drpeyam]

Precisely!!

[@Convergant]

This is some omega brain shit lmao

[@idkwhattosayxD]

Appreciate this approach! Thanks for posting the video

[@_schnelli4800]

Can't believe this is my first time hearing this trick despite how obvious it is, will definitely look to apply it from now on.

[@bishakhaalva800]

Who's here for CBSE class 12th boards 2024?

[@user-po7lx4vj6g]

broooo me after struggling from integration

[@TheQuantumArk]

this trick is awesome, as is your presentation!! loved the video

[@alaraf109]

Just impressed!!! I have learned to integrate ln(x), arcsin(x) etc and looking for arctan(x)...... And found your method!!! Can't express how happy I have become!!! 😁😁

[@whilewecan]

Incredible. I never thought so.... extremely good idea. Thank you.

[@jusinocasino11]

Waiiiittt why haven't anyone thought of this, It's simple yet genius. Thank you so much Dr Peyam

[@saloni.sharma]

what do i say except THANK YOU! and i absolutely love your energy! makes one want to study math with enthusiasm. :) this was very helpful indeed and i wish my teachers taught this way. all they did was give us a sheet of results they wanted us to memorise, explain first few of them and then give practice problems. :/

[@demo_yt5495]

absolute game changer!! Thank you.

[@solsticeyugen]

You have helped me achieve a new awareness in life. Thank you.

[@ruptyx242]

This is such a powerful clever trick!

[@socraticmathtutor1869]

Wonderful! Wish I'd known about this technique sooner.

[@misterdubity3073]

Wow!! I did not see that coming. Very powerful trick. And I like the "marker toss" at the end. "Marker drop" will be the new "mic drop". Subscribing now.

[@astonallsopp5746]

I love his enthusiasm

[@Cynic370]

Holy cheeze!, yesterday I was struggling with this integral, and I crunched it so many times without getting any easier expresion... And the answer was just there!

[@shourya9998]

This video came in my feed many times, glad I finally watched it. Love you Dr peyam!

[@drpeyam]

Thank you!!!!

[@lifeofsahani]

This is truly life changing!!!

[@omingnus]

Wow. Genuinely impressive.

[@marcinleonkempka2684]

This was great! Thank you!

[@zackwhite6199]

This video truly lives up to its title.

[@sk3105]

nahhh this is actually insane..my life feels like a lie for putting so much effort and not learning this earlier..thanks a ton for this

[@shelumieldilen]

Wow. This is truly awesome.