@Egad well that seems so obvious that we only write the simplest one (choosing the constant to be zero), because we're not asked to solve for the set of primitives but for a primitive. Also it's simply redundant, especially when it comes to partial derivatives (heat equation, wave propagation equation, or just differential calculus in math)
Nice! One suggestion: At 7:40 you have the integral of (x - 2)/((x - 1/2)^2 + 3/4) and then you split the numerator into x and -2. But it's easy to see that you will have to make a substitution for x - 1/2 later, so you can save some work by instead splitting the numerator into x - 1/2 and -3/2.
This works but it's faster to multiply the integrand by 2 before completing the square, and put 1/2 on the outside. This allows us to break the numerator into 2x-1 and -3, which lets us split the fraction into two with those numerators, then integrate the (2x-1)/x^2-x+1 via ln, and complete the square with -3/x^2-x+1 to get an easy arctan
Yeah, I did that. It makes that part a whole lot easier. Tbh, I recently discovered the difference of two cubes formula and I can prove it too, so I factorised, went to partial fractions, and solved it thereafter. I do love this integral. It is an interesting one.
At 2:09 you can directly substitute x = -1 to find the value of A = 1/3 The main idea is to directly find the value of (Bx + C) as follows as finding individual terms are not required: Substitute at 2:09 x2 - x = -1 from which we get Bx + C = 1/(x + 1); then multiply both num and deno by x + k Now to find the value of k equate the coefficient of x term of x2 - x and x term of x2 + (k+1)x + k from which k = -2 and Bx + C = -(x-2)/3
I remember doing this one as a definite integral from 1 to infinity. After the fraction decomposition, I did a bunch of weird splitting and manipulating of the integral, in which I ended up with two integrals that were divergent, plus some other stuff. However, the two divergent integrals could be combined to make something convergent; I also needed to do a little l’hopital, but I got there in the long run.
By the way, Euler found a general solution for that integral (from 0 to inf) i.e. for Int 1 / (x^n + 1) = π / (n sin (π/n) ), where n > 0, so your boi from 0 to inf evaluates to 2π / 3√3.
All I would do differently was from 6:40 , instead of completing the square - i would manipulate the numerator to be 2x-1 so u-sub can be used ; which would leave you with just a constant on the numerator from where you could do completing the square +trig sub
I'm Japanese student. I took this topic when I'm senior in high school. At the time, I have been studying it, using textbook in mathematics. The textbook is written about the way to solve the mathematics problem, combining various field of mathematics such as trigonometric function, the Integral and so on.
I just integrated this today and this came up at my recommended (UA-cam spying on me probably lol). It is faster to integrate it if at the second fraction after pfd you set u=denominator, then du=(2x-1)dx. Then you can break the numerator to 1/2*(2x-1 -3). Split it into 1/2(2x-1) and - 3/2. After that you split the integral and you have integral of 1/u du and another one which evaluates to a single arctan. Also, in the end inside the second ln the expression is always positive so you can remove the absolute value
Bravo! I checked your answer it is correct. There is a very useful rule which you can apply to make your life much easier. Let me give you an example. ∫(2 x-1)/((x^(2 )-x+1)) = l n |x^(2 )-x+1| when you have derivative of the denominator in the numerator you do not have to substitute. You just take the log. You were having X in the numerator, you could change it to the required derivative easily. there are too many substitutions. These could have been avoided. I am sure you will learn with experience.
How would you generalize this as the integral from 0 to 1 of 1/(1+t^x) written as a function with respect to t? If you defined a function f(x) as the above, think about this: f(0)=1/2 f(1)=ln(2) f(2)=π/4 Then if you plugged in f(3), it just becomes a big mess.
Yeah, you can take advantage of the fact that any odd polynomial has at least one real root to reduce it by a degree. In some cases you get lucky and the remaining non-linear factor may be further reducible. At the least, it's an easier problem when you split it using partial fraction decomposition. I love how cleanly Flammable writes and presents the material, though!
You used linearity of integral too early 8:13 Better substitution woulid be x-\frac{1}{2}=\frac{\sqrt{3}}{2}u Answer is correct and i dont see any mistakes , but it is not the fastest way Could you calculate Int{\frac{dx}{x^2(4x^2-3)^2\sqrt{x^2-1}}} with two different substitution (first Euler's substitution and inverse trig substitution) This will show that not always trig substitution is the fastest
Main reason I did it was to simplify the answer while integrating cube root of tangent x. I did t=cbrt(tan(x)) and after that u=t^2. Could you do that integral?
curious what the exclamation on the equals is for, I know it isn't significant, similarly we write a question mark when we 're not sure if equality holds. So does exclamation denote a revelation? :D
You could shorten the last bit a little by instead of splitting the x-2 numerator into two fractions, you could do .5(2x-1-3) and first integral would have numerator of .5(2x-1) while the second gas .5(-3). It just cuts down on an integral
Really nice job. It's been awhile since I've done integration by parts. The only thing I would avoid is saying "times dx". Even though we often treat it as multiplication for substitution purposes that's not really what's happening. It's more accurate to say "in terms of x". Similar thinking to why we say dy/dx isn't division.
Brian Purcell dy/dx is à division. dy/dx is the limit of a quotient, so by définition it is a quotient. But dy and dx are infitesimals which mean they can be as small as we want. And if you take physics courses it will help you to understand the meaning of partial derivatives when we evaluate the gradient of a function.
As far as I know dy/dx is division .. It is the slope of the function at a certain point which by definition is the change in y divided by the change in x
How about the integration of square root of (x^3 +1)dx. If you have some an idea how to solve it in shorter way or in longer way. I tried answering this one but I dont really sure if its correct.
Hi again :) did you already seen dr peyam's video about these type of integrals? You could check it out for the comprobation. Warning :v involucrates complex analysis
🙌🙏 I must humbly ask, why would you do something that anyone else could do when given the opertunity to do the opposite. I don't watch blackpenredpen anymore because he's basically a human Wolfram alpha. Your meme videos are why I subscribed, and I patiently await an other addition. In return I will help you with your new camera 💱
I also noticed you don't pronounce thorn or delta. Is there a reason why? I thought Germanic languages fathered these sounds? Or was it the other way around... Hm..."one (s)ird""one þird"=1/3; "(d)is""δis"=this
@@PapaFlammy69 Watching you going through this integral makes me understand how my high school students feels about the introductory examples I used when demonstrating substitution/change of variables last week.
Me: I got it! I got it!
P.F.: "plus some arbitrary constant."
Me: @#%&!!!!
In France we don't need to add the arbitrary constant, strange
@@whatever-td1nh well you should need it, or you don't solve for all the primitive functions
@Egad well that seems so obvious that we only write the simplest one (choosing the constant to be zero), because we're not asked to solve for the set of primitives but for a primitive. Also it's simply redundant, especially when it comes to partial derivatives (heat equation, wave propagation equation, or just differential calculus in math)
Blah & Bleh. My two new favourite variables xD
better than foo and bar
integration is an art
So that it's geometrical interpretation is area under the curve and two absissa😁😁😁
Integration is a sorcery
Lora mera
Is a nightmare.
Nice! One suggestion: At 7:40 you have the integral of (x - 2)/((x - 1/2)^2 + 3/4) and then you split the numerator into x and -2. But it's easy to see that you will have to make a substitution for x - 1/2 later, so you can save some work by instead splitting the numerator into x - 1/2 and -3/2.
I agree, Martin Epstein. flammable math's method is way too cumbersome.
This works but it's faster to multiply the integrand by 2 before completing the square, and put 1/2 on the outside. This allows us to break the numerator into 2x-1 and -3, which lets us split the fraction into two with those numerators, then integrate the (2x-1)/x^2-x+1 via ln, and complete the square with -3/x^2-x+1 to get an easy arctan
This has become one of my favorite videos. I spent like two hours doing that integral by other methods and you have taught me a nice one. Thank you.
Feels so weird seeing Papa Flammy not being as flashy as he is these days on camera... goes to show how far he's gone as a UA-camr.
For the Partial Fractions, you could just substitute x=-1 and that would remove (Bx+C)(x+1) and leaves us with 1=3A
Yeah, I did that. It makes that part a whole lot easier. Tbh, I recently discovered the difference of two cubes formula and I can prove it too, so I factorised, went to partial fractions, and solved it thereafter. I do love this integral. It is an interesting one.
That integral is SAVAGE!
so differentiate the answer and see if fits ;)
Left as an exercise to the viewer
Impressed with how you explain partial fraction decomposition.Find it really helpful.
At 2:09 you can directly substitute x = -1 to find the value of A = 1/3
The main idea is to directly find the value of (Bx + C) as follows as finding individual terms are not required:
Substitute at 2:09 x2 - x = -1 from which we get Bx + C = 1/(x + 1); then multiply both num and deno by x + k
Now to find the value of k equate the coefficient of x term of x2 - x and x term of x2 + (k+1)x + k from which k = -2
and Bx + C = -(x-2)/3
I like the not-complex-analysis version. Though I must ask if you could/have integrate(d) ln(x)? And if you already have, instead ln(1/ln(x))
I remember doing this one as a definite integral from 1 to infinity. After the fraction decomposition, I did a bunch of weird splitting and manipulating of the integral, in which I ended up with two integrals that were divergent, plus some other stuff. However, the two divergent integrals could be combined to make something convergent; I also needed to do a little l’hopital, but I got there in the long run.
The absolute value sign of the(1/6)ln|x²-x+1| in the final answer is no need to be added as x²-x+1is always positive
By the way, Euler found a general solution for that integral (from 0 to inf) i.e. for Int 1 / (x^n + 1) = π / (n sin (π/n) ), where n > 0, so your boi from 0 to inf evaluates to 2π / 3√3.
FM proved it too in one of his videos.
All I would do differently was from 6:40 , instead of completing the square - i would manipulate the numerator to be 2x-1 so u-sub can be used ; which would leave you with just a constant on the numerator from where you could do completing the square +trig sub
You’re a real life superhero
I'm Japanese student. I took this topic when I'm senior in high school. At the time, I have been studying it, using textbook in mathematics. The textbook is written about the way to solve the mathematics problem, combining various field of mathematics such as trigonometric function, the Integral and so on.
I just integrated this today and this came up at my recommended (UA-cam spying on me probably lol). It is faster to integrate it if at the second fraction after pfd you set u=denominator, then du=(2x-1)dx. Then you can break the numerator to 1/2*(2x-1 -3). Split it into 1/2(2x-1) and - 3/2. After that you split the integral and you have integral of 1/u du and another one which evaluates to a single arctan. Also, in the end inside the second ln the expression is always positive so you can remove the absolute value
At 6:30 let u=x^2-x to get rid of that nasty boi. Still left with a hot piece tho.
You really saved me man, thanks a lot!
Greetings from a savage engineer :)
this is beautiful
Ich habe dich abonniert! 😊
Grüße aus England!
Flammable Maths Gern geschehen :)
Bravo! I checked your answer it is correct.
There is a very useful rule which you can apply to make your life much easier. Let me give you an example.
∫(2 x-1)/((x^(2 )-x+1)) = l n |x^(2 )-x+1| when you have derivative of the denominator in the numerator you do not have to substitute. You just take the log. You were having X in the numerator, you could change it to the required derivative easily. there are too many substitutions. These could have been avoided. I am sure you will learn with experience.
That's really helpful and easy!! Integrals are my favorite from now😀
When you’ve studied integration for so long you can plan out the full solution to this problem in your head
Excellent.
How would you generalize this as the integral from 0 to 1 of 1/(1+t^x) written as a function with respect to t?
If you defined a function f(x) as the above, think about this:
f(0)=1/2
f(1)=ln(2)
f(2)=π/4
Then if you plugged in f(3), it just becomes a big mess.
Thanks for this explanation, I like your style too
Let g be a continuous function which is not differentiable at 0 and let g(0) = 8. If
f(x) = xg(x), then f '(0)=?
8😣
Oh so nice trick to solve cubic equation integration
Yeah, you can take advantage of the fact that any odd polynomial has at least one real root to reduce it by a degree. In some cases you get lucky and the remaining non-linear factor may be further reducible. At the least, it's an easier problem when you split it using partial fraction decomposition. I love how cleanly Flammable writes and presents the material, though!
Why did the algebraic term 4/(3√3) become 1/√3?
Wow, I must be the busy little homunculus inside of Mathematica. I hope WolframAlpha is paying for this!-)
You used linearity of integral too early
8:13 Better substitution woulid be x-\frac{1}{2}=\frac{\sqrt{3}}{2}u
Answer is correct and i dont see any mistakes , but it is not the fastest way
Could you calculate Int{\frac{dx}{x^2(4x^2-3)^2\sqrt{x^2-1}}}
with two different substitution (first Euler's substitution and inverse trig substitution)
This will show that not always trig substitution is the fastest
Excellent professor.
I just learned something today.
I've gone further and the logarithm part I have is: 1/6 ln((x+1)^3/(x^3+1))
Main reason I did it was to simplify the answer while integrating cube root of tangent x. I did t=cbrt(tan(x)) and after that u=t^2. Could you do that integral?
curious what the exclamation on the equals is for, I know it isn't significant, similarly we write a question mark when we 're not sure if equality holds. So does exclamation denote a revelation? :D
I came for the math and stayed for the voice.
I think there's a problem at 12:19, there's a v you completely forgot to substitute and therefore i assume the final integral is not correct
I just solved this question in my maths test today the same way , I was finding more efficient way to solve this but can't find any
You could shorten the last bit a little by instead of splitting the x-2 numerator into two fractions, you could do .5(2x-1-3) and first integral would have numerator of .5(2x-1) while the second gas .5(-3). It just cuts down on an integral
Which class are you in?
Welcher Bahnhof ist das am Ende? kommt mir bekannt vor
This boi is good !
failed in my 12th grade math, 8 years later I woke up realising that I'm quite good at math
Well done👍🏻
"I can't believe this came out on my birthday. What a gift." - Jesus
Superb solution
Welcome sir. I also love calculus
I once done 1/(x^5+1) to prove some guy that was hard. It is easy in concept but hard with numbers
That wasn’t so hard. You could split the x-2 into (2x-1-3)/2 and with substitutions, you get the typical log and arctan functions.
wow thanks a lot, teacher
thanks a lot!❣️
thanks a lot.
Love from India ❤️❤️
Wow that was awesome
Omg :o , Good video!
Really nice job. It's been awhile since I've done integration by parts. The only thing I would avoid is saying "times dx". Even though we often treat it as multiplication for substitution purposes that's not really what's happening. It's more accurate to say "in terms of x". Similar thinking to why we say dy/dx isn't division.
Brian Purcell dy/dx is à division. dy/dx is the limit of a quotient, so by définition it is a quotient. But dy and dx are infitesimals which mean they can be as small as we want. And if you take physics courses it will help you to understand the meaning of partial derivatives when we evaluate the gradient of a function.
As far as I know dy/dx is division .. It is the slope of the function at a certain point which by definition is the change in y divided by the change in x
I've actually done 1/(x^6+1) by hand.
Not pretty.
Unless you like trig sub and about 4 years of cancellation, rearranging, and indentities.
How about the integration of square root of (x^3 +1)dx. If you have some an idea how to solve it in shorter way or in longer way. I tried answering this one but I dont really sure if its correct.
Cannot see the writing on the board.
Thank you sir 😃😃😃♥️♥️♥️
Brother integrate this one sqrt((7x²+8x+9)/(10x²+11x+12))
Hi again :) did you already seen dr peyam's video about these type of integrals? You could check it out for the comprobation. Warning :v involucrates complex analysis
Integrate 1/(sinx)^6+(cosx)^6
Hiiiiiii , j'arrive pas à comprendre comment ça se fait en arctan , comment on a fait sortir la racine carré de trois demi , s'il vous plaît aidez moi
Omg you really help me,I think I love you...hahahhha
🙌🙏 I must humbly ask, why would you do something that anyone else could do when given the opertunity to do the opposite. I don't watch blackpenredpen anymore because he's basically a human Wolfram alpha. Your meme videos are why I subscribed, and I patiently await an other addition. In return I will help you with your new camera 💱
Me acabas de salvar la tarea 😏😏😏
can some one explain me why i need in the numeratior degree - 1 of denminator when im doing PFD
any higher just cancels out^^
@@PapaFlammy69 and Lower degree?
@@Oskar-zt9dc The lower degree is already possible, if the numerator is Ax+B for example, you could just get A=0.
A real man would evaluate ∫ 1/(1 + x⁵)dx.
Thanks sir..
good teacher where do you live
Nice video. Flammable Maths sounds better, like you're on fire!
How can I integrate 2/(1+3x^3)? Plz brother help me in this.
Hint: u = 3^(-3) x
@@Xrelent I think you mean 3^(1/3)
@@98danielray oops, yeah you're right
First semester in engineering be like
Didjtn anyone else do it by integration by parts starting with ln of (x^3+1)??
differentiate it back please :)
I also noticed you don't pronounce thorn or delta. Is there a reason why? I thought Germanic languages fathered these sounds? Or was it the other way around... Hm..."one (s)ird""one þird"=1/3; "(d)is""δis"=this
Ryan Roberson *one (f)ird
x^2 cancels out only because x^2 cannot be zero
Mr Beast: say "ln" and win $80.000
Chandler: 14:24
3:46 shouldn't it be +ax not -ax
Avi Mehra Nope, A = - B
Finally, as a Physics major I can calculate something that is given from our Lord and Savior papa Flammable maths. XD
Ok but, what that answer mean?
I easily understand it
I had it on my first math exam at uni :/
🙌
Oh dude thank you is usefull to ! #Flame
pretty extravagant though :D
definitely! :D
Just do the geometric series
Yay!
Maths lovers like here💖
Thnks
It's very lengthy and time consuming
all the boards you use are sliding up lol
not very interesting integral to be honest. Something you would find in a textbook or smth
(x-2)/(x^2-x+1) = 1/2[(2x-1)/(x^2-x+1) - 1/(x^2-x+1)] this may have eased your job
You’ve missed out a -1/(x^2-x+1) though.
It’s trivial, but it’s so fucking long
Actually hate integrals like this, tedious!!
Okay.
@@PapaFlammy69 Watching you going through this integral makes me understand how my high school students feels about the introductory examples I used when demonstrating substitution/change of variables last week.
oof