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.
@@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.
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.
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!
@@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
@@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
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
@@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?"
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.
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 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.
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).
@@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!
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
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!!! 😁😁
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 ^^
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)?
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.
OHH GODD! I HAVE BEEN TRANSFORMED AND WILL NEVER LOOK AT ANY INTEGRAL THE SAME WAY AS I DID EARLIER!!! This is just ingenious and astoundingly simply, yet so effective!
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
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
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!
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.
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
@@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
@@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.
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.
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
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
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!
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.
0:46 I use an add-and-subtract method here; (x²)/(x²+1) = (x² + 1 - 1)/(x² + 1) = (x² + 1)/(x² + 1) - 1/(x² + 1) = 1 - 1/(x² + 1); this becomes much easier to integrate It's just like doing polynomial long division, except you don't cry while dividing
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.
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)"
the best method is DI method where we make a small table and write d and i on top which refers to differentiation and integration. then chose the first function and second function and the first function is always differentiate so keep the first function on the D column and the second on I column. suppose we take x^2 and sinx and x^2 i put it in D column and sinx in I. now simply keep on differentiating till you get zero in first column and keep on integrating the second . now change the signs alternatively on D column (start with + then go with - and then + and so on)anddd cross multiply from D column and VOILA YOU GOT IT :)
@@drpeyamAlright I made it to Calc 2 and my god this trick is hilarious my brain wants to think this way but years of schooling makes me doubt thinking like this.
My mind is blown. Why is this mot taught?! Why have I literally never come across this amaizng trick in any textbooks, in any exams, by any teachers?! This is something that should be widely known and taught!!!
Very useful trick i wish every maths content creator should come up with trick like that to make the life of students easy and also it creates interest in maths
This sort of reminds me of 7th grade, when we learned about "completing the square." This trick is really smart and I will use it in the coming years. Thank you :)
Clever trick! I didn't understand at first why the trick is legit. This integral is solved by using the rule ∫u dv = uv - ∫v du. Here, dv is x dx. When integrating dv, we obtain x²/2 to which we can add a constant C.
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.
lmao
but, be careful, if you watch the video backwards, it turns into a Country Western music video…
LMAOOO
LMAO
LMAOOOOOO
Now the +C is actually useful. Great video
Yup
So trueee
This is actually really cool 😆
Glad I actually watched it haha
Awwwww thank youuuu!!!
Please combine this with the DI method!😺
Nuuuuu! Now bprp will kill his students with these questions!
Master 😅
What is this crossover episode? 🧐
How did I not know this. My life is changed.
Greetings sir
Orz here
It's a very normal question in WBCHSE class 12
@@abhwiNo one teaches this method in class 12
same here!!!!!
I've taught this method to my students for years and it's always made them much better at integrating in general
Hey, I have a doubt, Why do we have to choose the same constant at both the places though? for example at 4:47
symbolab gave me another result in his first integral
A Arjith look at the proof of integration by parts
@@aarjith2580 because we can't choose different ones. We are plugging in the antiderivative in both places you bum
@@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.
"Arent you glad I found my video? 😁" Yes! So happy!
Glad to see you so happy about math too :)
3:11 “Isnt that nice!” Nice ? NICEEE ??? SIR THAT IS STRAIGHT AMAZING my life is never the same after this
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.
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!
I'm sure it was more a result of you asking your question unclearly
Probably your question wasn’t clear to him, so he meant you don’t add +C within each term.
@@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
I think he meant you were asking if you can leave constants in the equation without absorbing them into +C
Thought the title was click bait.
It was NOT click bait.
Life changing, indeed!
Omg hi Floyd!!!
@@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
@@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
I love his attitude, man knows math and is happy about teaching it. I wish all my teachers were like him.
As a physic student I can say, this is really lifechnaging, hours of pain will be joy instead. Thanks 🙏🏼
Wow, such a simple idea that's hidden in plain sight! Just in time for my calculus test this week.
Good luck for your exam bro.
blessed to have him as my differential equations professor, couldn’t be luckier
Awwwww thank you ❤️ It’s symmetric!
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
Yeah I ve thought the same thing
Me 2
yeah i agree but this does seem to be a nice tip
Nope. The 'trick' only works in cases where the algebraic simplification also works. This is not a method, it is nothing.
@@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?"
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.
Well, it is a function if you consider an equivalence relation between functions that differ by a constant.
(And then it is a linear isomorphism)
👆 put differently, Give it a couple years and you'll stop thinking of constants as really making them "different"
I've learnt the +1 -1 trick. on integration but this takes it to a new level - can't thank you enough!
Me too
How have I never seen this before? This is amazing!
Omg hi Mu Prime Math !!!
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!
int (1+x+x^2)/(1+x) dx = int 1 + x^2/(1+x) dx, then use u = 1+x.
@@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.
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).
@@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!
@@violintegral noicely done bro ... Yep we can use substitution.. but don't forget dy 😉
This blew my mind! So fascinating. I guess that's why in my courses it usually says "find the most 'general' antiderivative."
Wow, I expected this to be exaggerated, but it's actually really cool
For the ln(x+2)dx integral, you can substitute u=x+2 as well to bring it to a standard form
He is using the same method but not showing you the substitution. d(x+C) where C is a constant
Here's an example with ln(), that is more interesting.
integral ln(|x^2 - 4|) dx
I gotta say I love learning math from people who talk about its tricks and properties with such passion. Thank you for your video!!
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
I never even considered this. Makes it so much easier!
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
Yep this would also make the ln(x+2) integral easier too
@@robertveith6383 The area under 1.x yes, but simply ln(x) for integral, thats what Wolfram Alpha says anyhow.
It's inspired from Ross's analysis book.
isn't that just substitution?
@@coerciasink Indeed but works in general for any function f(x+c) -> f(u)
This trick is amazingly amazing, this really changed my life and how I see Integrals, thank you Dr. Peyam!
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 ❤️❤️❤️❤️
It s amazing how something so important and so simple ,isn't used often
Just can't explain how grateful I am to you for explaining this miraculously magical method
Thanks alot ❤️
I can't express how glad I am after getting to know this trick as an Aspiring Engineer ! Thank you Dr Peyam
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 😁
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!!! 😁😁
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 ^^
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)?
This video came in my feed many times, glad I finally watched it.
Love you Dr peyam!
Thank you!!!!
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.
Sorry legend
Yea its easy
Smartass
That is mind blowing, I used it on my math test and the integral went so smooth, thanks from the deepest part of my heart
Omg so glad it was useful!!
Absolutely stellar expedient method!
OHH GODD! I HAVE BEEN TRANSFORMED AND WILL NEVER LOOK AT ANY INTEGRAL THE SAME WAY AS I DID EARLIER!!!
This is just ingenious and astoundingly simply, yet so effective!
@@robertveith6383 OK
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
Yes, by adding a zero, you can obtain the same result as the trick although you might need to do a couple more steps.
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.
Mann that boom boom 3:01 !!! was so energetic 😅🎉
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
this is some serious math hack territory we are treading in
Who knew adding one could be so revolutionary
Life changing man, for real 💯
Nice!!!
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!
I don't know why, but this makes me really happy 🤤🤗🖤🖤
Me 2😚
Solving integrals is allways fun!!!
My lord, such a simple yet elegant trick and it's so convenient.
Many thanks for sharing it with us!
Truly exciting and imaginative!
I have no idea how this ended up in my recommended feed, but THANK GOD IT DID
What an amazing trick !!!
Thank you for sharing.
This video truly lives up to its title.
Solving integrals is allways fun!!!
@Dr Peyam What a shock...a pleasant shock. Integration by parts will never be the same again for me
This feels illegal
i absolutely despise by parts, but this changed my life! thanks!
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.
Why not? Any antiderivative is equally legit, isn't it?
Of course! When you choose the antiderivative in DI, just pick the right constant
@@drpeyam yah
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.
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
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
This is a God like power for integration !
This is so awesome!
totally agree
Waiiiittt why haven't anyone thought of this, It's simple yet genius. Thank you so much Dr Peyam
Being a JEE aspirant , its a very normal technique taught to us. However, its really nice of you to educate people so passionately.
You're the man!! Usually it takes more than one videos to get me to subscribe.
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
How do you figure out which integrals it does/doesn’t work for?
@@uwuifyingransomware im guessing by practice
@@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
@@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.
I literally have a calculus 2 exam tomorrow, this makes it so much easier
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.
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
Beautiful tricks never cross one's mind until someone with enough insights reveal them. Thank you Dr. Peyam!
I don't even know integration 😂
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
Wow. Genuinely impressive.
I can't believe my eyes. Why didnt we know this before?
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!
5:13
My favorite part of this video 😎
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.
Dr. Peyam: Thanks for watching.
Me: Thanks for changing my life!
Incredible. Thank you so much for this trick!
0:01 "Thanks for Watching". 😂😂
0:46 I use an add-and-subtract method here;
(x²)/(x²+1) = (x² + 1 - 1)/(x² + 1) = (x² + 1)/(x² + 1) - 1/(x² + 1) = 1 - 1/(x² + 1); this becomes much easier to integrate
It's just like doing polynomial long division, except you don't cry while dividing
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.
a neat concept that clarifies the +c as well👍👍👍
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)"
Precisely!!
this appeared on my feed when I was procrastinating my homework in calculus
This is some omega brain shit lmao
the best method is DI method where we make a small table and write d and i on top which refers to differentiation and integration. then chose the first function and second function and the first function is always differentiate so keep the first function on the D column and the second on I column. suppose we take x^2 and sinx and x^2 i put it in D column and sinx in I. now simply keep on differentiating till you get zero in first column and keep on integrating the second . now change the signs alternatively on D column (start with + then go with - and then + and so on)anddd cross multiply from D column and VOILA YOU GOT IT :)
Who's here for CBSE class 12th boards 2024?
broooo me after struggling from integration
This video had so much charisma, I can’t believe it. Made me lol at the relax comment!
Glad you enjoyed it!
@@drpeyamAlright I made it to Calc 2 and my god this trick is hilarious my brain wants to think this way but years of schooling makes me doubt thinking like this.
My mind is blown. Why is this mot taught?! Why have I literally never come across this amaizng trick in any textbooks, in any exams, by any teachers?! This is something that should be widely known and taught!!!
Thank you!!!!
I found this when I’ve been suffering the most with Calc 2 thank you kind sir
this would’ve been helpful last semester. thanks, youtube recommended.
Very useful trick i wish every maths content creator should come up with trick like that to make the life of students easy and also it creates interest in maths
That was actually incredible.
BRO this helps me so much with speeding up my solving time omg
thank you so so much, you're the goat frfr
You’re welcome :)
This sort of reminds me of 7th grade, when we learned about "completing the square."
This trick is really smart and I will use it in the coming years. Thank you :)
Truly life-changing. Was absolutely mind-blown
Im almost crying how good this is , Thank you very much for sharing
'+C' is the most underrated part of antiderivatives.
Never thought about this before!
Clever trick! I didn't understand at first why the trick is legit. This integral is solved by using the rule ∫u dv = uv - ∫v du. Here, dv is x dx. When integrating dv, we obtain x²/2 to which we can add a constant C.
Totally blown my mind! Shared to all my calc homies
Please make more such videos! Thanks!
This is mind-blowing I love it, never thought of taking advantage of the constant like this. Nice!!