I have a extremely easy method which use hyperbolic functions to determine the value of this at certain limits or even indefinite integral. Are you interested ?(you say i dont care about your solution)_(that hurts)😪
I love how he remade a 25ish minute video just because of that small mistake. It shows his dedication. I love how right after he gets to his past mistake he starts acting more calm like he’s finally redeemed himself.
I was wondering if he intentionally sped the video up to impress us, so I slowed it down to .75 speed. At that speed, though, he sounds drunk, so he must really be talking that fast. Dang...
@@zanti4132 There are two ways to speed up or slow down the audio. RUclips uses the spectrogram to speed up and slow down audio, resulting in no pitch shift, but if you speed up some audio as it is usually inputted, stored, and outputted, the pitch gets higher, and vice versa
Agh, I know, it's amazing how enthusiastic you two are. I was talking to one of my friends the other day and the only word I can really use to describe it is that your guys' energy is just so infectious; watching your guys' videos is enough to just make me smile, it's absolutely amazing. Enthusiasm makes all of the difference - my numerical methods professor always looks like she wants to go be anywhere but there in our class and doesn't really have ANY energy so I always feel like going to sleep in her class, but then I see you guys, or my foundations of math prof or my number theory prof and all of you are so lighthearted that I can actually enjoy myself. Energy and passion really make quite a difference on imparting material to people from an educator's standpoint.
@@blackpenredpen Why didnt you do partial fractions by factoring out the x^2 instead im curious? To get a product of 1/x^2(x^2 + 1/x^2)? That works too..
Another method to solve this is to multiply numerator and denominator by 1/(4x^2). Then breakfast the numerator as {1+1/(2x^2} + {1-1/(2x^2)}. This method will not require partial fractions.
Thank You so much for this video. I know that making videos like this takes up your time and i'm so grateful that there are people like You that have the patience and the good will to help people like me. :D
You are the most amazing mathematician on the internet. I can't thank you enough for the fantastic instruction you offer and the help you provide. The fluidity and speed with which you are capable of doing extremely complex problems is the most enviable thing I've ever seen. You are flat out AMAZING.
Thank you for all these videos. I really enjoy watching them. I have always loved math, became an engineer, and had not seen these problems in a long time. Good to refresh.
OMG, when I saw that system of equations thought it was going to take a very long time and be hard, but you're way of handling it is just genius. It was so fast. Wow...
blackpenredpen I wouldn't want to go through that mess to find the integral over R of (1+x^4)^(-1) if I knew i could just use contour integration and get the result way more easily
I honestly thought that was more a part of his whole "I'mma do it my way not your way" thing. Which is all good: so long as the steps and result are justified and make sense, anything goes.
I remember your comment Bosk. That's why I laughed. Plus, I wouldn't use the word "trigger". Since I usually find it really laughable (instead of getting mad or triggered) when people comment on my vids about what to do. Thus, thanks for making me laugh, twice. That also gave me the idea to say the line that Daniel mentioned. Btw, it's all good. Some of my viewers are 12 anyway. : )
Very nice! It's interesting that you can get a formula for the antiderivative of this function. A lot of rational functions like this can't be integrated by normal means and require complex analysis to compute.
This channel has the best meth videos on the internet. (Your accent gets a lot stronger when you speak that fast + it already requires quite a lot of mental processing to compensate for a Chinese accent if one isn't used to it = aaargh!)
Maybe it's just me but I didn't have a problem understanding his accent (I am in mid-US) although, since it was a bit rushed, I need to listen through a couple of times to digest the flood of information. However, I'd have to go this fast in an exam, so...
Not a lot of English speaking Chinese in Continental Europe... Not on TV, either. A Chinese accent means almost all word endings (and syllable endings) get cut off -- which is problematic for Indo European languages because that's where all the inflections are. He did okay in another video where he spoke slower. He did that video in Chinese, too, which I surprisingly understood a bit of. The first part of this video was harder for me to understand than his video in Chinese! (I used the Hello Chinese app to learn some very basic Chinese.)
I'd have to specifically explore it, but the two natural logarithms can be combined, and the two arctangents can also be combined although there are minor problems with respect to x-values.
I had to do this exact integral as uni homework last week. Let's just say it didn't go well, but I was so frustrated that I couldn't solve it, that I spent two hours with wolfram alpha yesterday trying to understand the solution. What I still don't get is how I know which approach to take. It feels like there are so many different ways to go after the partial fraction with this integral.
1/(1+x^4)=1/x^2/(x^2+1/x^2) =(1+1/x^2)/((x-1/x)^2+2)-(1-1/x^2)/((x-1/x)^2+2) Now put u=(1-1/x) and v=(1+1/x) Integration can be easily done in terms of u and v.
This is my comment so I can write whatever I want to write !! Awesome video - understood everything - saved me some time because I had to finish plenty more pages and I had this for certain residue theorem . Thanks bro !!
You can take x^2 common in the first step. Then you multiply and divide by 2 . Then add and substract 1 in the numerator . And so you have (x+1/x)^2 - 2 in the denominator and 1-1/x^2 in the numerator and and the integral becomes that of arctanx. And for the second part you have (x-1/x)^2 +2 in the denominator and 1+1/x^2 in the numerator/ use substitution and it becomes a partial fraction and you can integrate to give a logarithmic function . There is no need to factorise the expression.
04:11 ~ 04:43 (from transcript) "and now let's combine all the x squared term namely we have this right here let's just put on the liner this until I got this and then this is another two right here right so we have this Plus this Plus that but that's that sorry this minus this and the plaster a plaster right so in total we have to a minus square root of 2 C and a plus B plus D oh this right here is the coefficient of x squared on the right hand side but once again we don't have x squared on the left hand side that means oh this has to be 0 that's good"
It could be done simpler: the numerator of the intetrand 1=1/2[(1+x^2) + (1-x^2)], then divide x^2 to both numerators and denominators before introducing new variables u=x-(1/x) for the first part integral and v=x+(1/x) for the second, they will end up with two results: one arctangent function and one logarithmic
We can do it in a simpler method, add subtract x² Seperate into 2 terms 1/2[(x²+1/x⁴+1)+(1-x²/x⁴+1)] Get x² out u get 1/2[(1+1/x²)/(x²+1/x²)+(1/x² - 1)/(x²+ 1/x²)] And reduce denominator to ( x+1/x)² -2 And take the numerator inside d (Substitute t=numerator and make it dt) And u get a form of {dt/t²-a²+ {dt/t²+a² Then u can apply the respective formula and get the answer
This was so hard! (Even though I was doing this, but at times, I was SO FAR BEHIND that I had to pause the clip just so I could catch up with you!) (BTW, I posted this from Iowa at 5:15 pm Central Time.)
Your work is fine in terms of getting a process that the integral is solvable. I appreciate it but, this problem has a very short and delicate solution with fewer steps of computation.
easier by factoring an x^2 from the denominator,and your integral becomes (1/x^2)/(x^2+1/x^2)dx than the denominator becomes either (x+1/x)^2-2 or (x-1/x)^2+2 and then force the derivative of the denominator to the nominator and done
Marcus Åkerman the derivative of (1/x+x) is - 1/x^2 +1 , and since you factored x^2 from the denominator you formed that derivative, and then you add and subtract 1 and you formed the integral in terms of itself and an integral which has the form of f'/f which is lnf
Write numerator= 1/2((x^2+1)-(x^2-1)) And break the denominator, you will get two standard forms of integrals and you will get your answer in 2 minutes with 2 basic steps
I decided to do this one on my own to see if I could do it and I ended up turning it into 1/2*integral(sqrt(cot(v))*dv) (where tan(v)=u and u=x^2) and I ended up having to watch your video on sqrt(tan(x)) I'm not sure which method is more difficult but I do like how the trig sub worked out (I also ended up using w, q, and r subs)
While the algebra is certainly interesting, this problem would become so much easier through the initial substitution x=t/SQRT{2} because then the denominator becomes t^4+4 which is perfectly factorable into the product of two quadratic polynomials. At the end you only have to backsub.
Fantastic video as usual! I have a (probably silly) question. How do you know, in the partial fractions, that the numerator is a first degree polynomial (Ax+B) and not any other degree? Thanks in advance!
Not a silly question at all! Notice that the denominators in the two parts of the partial fractions were quadratics, i.e. of degree two. That means that the numerator is exactly one degree less, so you need a degree one polynomial as the numerator. If the denominator happened to be a third degree polynomial, then you'd need a second degree one in the numerator, so something like Ax^2+Bx+C.
"I want to do it my way, I don't care about your way, it's my video right here!"
*- blackpenredpen*
Gameboygenius You should still watch the video in double speed.
Double speed is usually too slow, but no higher choice ...
THIS IS DONT THAT BAD AT ALL!!
Go int (1/(x^n+a^n))dx
I have a extremely easy method which use hyperbolic functions to determine the value of this at certain limits or even indefinite integral.
Are you interested ?(you say i dont care about your solution)_(that hurts)😪
I love how he remade a 25ish minute video just because of that small mistake. It shows his dedication. I love how right after he gets to his past mistake he starts acting more calm like he’s finally redeemed himself.
He not only became calmer, but he said (rightfully so) that he'd redeemed himself (PHEW).
You literally made me double check that I didn't accidentally speed up the video.
I was wondering if he intentionally sped the video up to impress us, so I slowed it down to .75 speed. At that speed, though, he sounds drunk, so he must really be talking that fast. Dang...
@@zanti4132 There are two ways to speed up or slow down the audio. RUclips uses the spectrogram to speed up and slow down audio, resulting in no pitch shift, but if you speed up some audio as it is usually inputted, stored, and outputted, the pitch gets higher, and vice versa
I love your enthusiasm for math. Why aren't more professors and teachers like that...
enverko thanks!! Dr. Peyam is definitely one as well.
Agh, I know, it's amazing how enthusiastic you two are. I was talking to one of my friends the other day and the only word I can really use to describe it is that your guys' energy is just so infectious; watching your guys' videos is enough to just make me smile, it's absolutely amazing. Enthusiasm makes all of the difference - my numerical methods professor always looks like she wants to go be anywhere but there in our class and doesn't really have ANY energy so I always feel like going to sleep in her class, but then I see you guys, or my foundations of math prof or my number theory prof and all of you are so lighthearted that I can actually enjoy myself. Energy and passion really make quite a difference on imparting material to people from an educator's standpoint.
@@blackpenredpen Why didnt you do partial fractions by factoring out the x^2 instead im curious? To get a product of 1/x^2(x^2 + 1/x^2)? That works too..
6:31 that “AT ALL” startled me
"I WANT TO DO IT MY WAY. I DONT CARE ABOUT YOUR WAY" LMAO
😂😂😂😂
Exactly 👍
Another method to solve this is to multiply numerator and denominator by 1/(4x^2). Then breakfast the numerator as {1+1/(2x^2} + {1-1/(2x^2)}. This method will not require partial fractions.
It's not fun if not
Thank You so much for this video. I know that making videos like this takes up your time and i'm so grateful that there are people like You that have the patience and the good will to help people like me. :D
"I'm going show you guys how to factor 1/(x^4+1)" *Initial D music starts playing*
neg atory DEJA VU
MATH MATH MATH (im gonna write all the math!)
😂😂
You are the most amazing mathematician on the internet. I can't thank you enough for the fantastic instruction you offer and the help you provide. The fluidity and speed with which you are capable of doing extremely complex problems is the most enviable thing I've ever seen. You are flat out AMAZING.
Horizon thank you!!
i think he is a genius. i was a tooter of calculus more than a half centuries . i easley say his iq is more than 160.
@@eng954Genius yes but 160 is a stretch, 140 is more plausible.
Thank you for all these videos. I really enjoy watching them. I have always loved math, became an engineer, and had not seen these problems in a long time. Good to refresh.
Still the best dressed mathematician on RUclips. Thanks for the awesome vids.
Nate Yerger thanks!!!
I laughed at the +C at the end, poor +C has been waiting for so long.
Too much coffee confirmed xD
Nope, *Big Asian Energy confirmed*
*methamphetamine
OMG, when I saw that system of equations thought it was going to take a very long time and be hard, but you're way of handling it is just genius. It was so fast. Wow...
This is precisely the reason we use complex analysis to evaluate integrals over R
The Physicist Cuber ?
blackpenredpen I wouldn't want to go through that mess to find the integral over R of (1+x^4)^(-1) if I knew i could just use contour integration and get the result way more easily
Well, you can't really find an indefinite form for the integral using contour integration, so this is still a necessary video.
The Physicist Cuber what if I just want the integral from 1 to 2 :)
blackpenredpen just wolfram xD
I no longer have the right to say I am good at math. Ohh lord.
It's only 2:32 am here, how about you?
11:39 (Europe)
What are you still doing so late ?! Go get some sleep !
It's 4:40 pm here
blackpenredpen 11:32 A.M. at the time you wrote the comment
Dracquiteur I wanted to get this done and uploaded! :)
The Physicist Cuber just now. Which is about 2:44am now.
It's 10:38 am here, also I like the supreme garms you wear in your videos
thanks!
5:47 "but I want to do it my way, I don't care about your way, and it's my video right here" made me laugh😂😂
"I want to do it my way i don't care about your way" so inspiring!!
awesome speed - and infectious enthusiasm. You make calculus a pleasure - thanks bprp for this awesome integral :)
that's equal in more compact form, if you check and prove, to:
=1/2√2 ( acoth 1/√2 (1/x+x) + acot 1/√2 (1/x-x) ) + C ;
"You guys kept telling me what to do, but I will do it my way because it's my video"....hilarious.
I felt going crazy and watched it in double speed. I'm a hero!
6:22 wasn't that a bit overkill?
You had:
√2B - √2D = 0
B + D = 1
Couldn't just set
√2B = √2D
B = D
2B = 1
B = D = 1/2
Bosk Boskson love it hahahahahaha
Haha :)
I honestly thought that was more a part of his whole "I'mma do it my way not your way" thing. Which is all good: so long as the steps and result are justified and make sense, anything goes.
I know, I just copy pasted my exact comment from the last video, just changed the time....Just to trigger him a bit ;)
I remember your comment Bosk. That's why I laughed.
Plus, I wouldn't use the word "trigger". Since I usually find it really laughable (instead of getting mad or triggered) when people comment on my vids about what to do. Thus, thanks for making me laugh, twice. That also gave me the idea to say the line that Daniel mentioned.
Btw, it's all good.
Some of my viewers are 12 anyway.
: )
Very nice! It's interesting that you can get a formula for the antiderivative of this function. A lot of rational functions like this can't be integrated by normal means and require complex analysis to compute.
Your way of teaching makes a person to listen . So good
I’m literally speechless...
You are literally the smartest math person I’ve ever encountered
Ok dumb albert
"AT ALL!"
This channel has the best meth videos on the internet.
(Your accent gets a lot stronger when you speak that fast + it already requires quite a lot of mental processing to compensate for a Chinese accent if one isn't used to it = aaargh!)
Maybe it's just me but I didn't have a problem understanding his accent (I am in mid-US) although, since it was a bit rushed, I need to listen through a couple of times to digest the flood of information. However, I'd have to go this fast in an exam, so...
Not a lot of English speaking Chinese in Continental Europe... Not on TV, either.
A Chinese accent means almost all word endings (and syllable endings) get cut off -- which is problematic for Indo European languages because that's where all the inflections are. He did okay in another video where he spoke slower. He did that video in Chinese, too, which I surprisingly understood a bit of. The first part of this video was harder for me to understand than his video in Chinese!
(I used the Hello Chinese app to learn some very basic Chinese.)
You can also factorize it as (×^2+i)*(x^2-i)
Massive respect for doing it from scratch again!
AcidicAlkali thanks!!!
Thanks a lot for understanding this question.
lots of love from india. keep solving such problems in your own ways.
¡¡No maa, neta eres de lo más chingón que pude haber , muchas gracias!!
I'd have to specifically explore it, but the two natural logarithms can be combined, and the two arctangents can also be combined although there are minor problems with respect to x-values.
I had to do this exact integral as uni homework last week. Let's just say it didn't go well, but I was so frustrated that I couldn't solve it, that I spent two hours with wolfram alpha yesterday trying to understand the solution.
What I still don't get is how I know which approach to take. It feels like there are so many different ways to go after the partial fraction with this integral.
i got this GODDAMN integral in my exam......and only after the exam YT is suggesting me this vid.....
This is how bprp fast began :)
This was the best thing today i saw in youtube.. it was simply too goood.. thnk u blackredpen
Watchout at 8:20 complete the square the correct result is not (1/sqrt(2))^2 but (1/2*sqrt(2))^2=(sqrt(2)/2)^2
Genius. ABSOLUTE GENIUS. That's what you are.
1/(1+x^4)=1/x^2/(x^2+1/x^2)
=(1+1/x^2)/((x-1/x)^2+2)-(1-1/x^2)/((x-1/x)^2+2)
Now put u=(1-1/x) and v=(1+1/x)
Integration can be easily done in terms of u and v.
This is my comment so I can write whatever I want to write !!
Awesome video - understood everything - saved me some time because I had to finish plenty more pages and I had this for certain residue theorem .
Thanks bro !!
You're welcome!
Imagine doing all of this to realise that you forgate a x on the top
Our calculus 2 teacher gave us this as a homework , thank u very much
5:46 "But I want to do it my way. I don't care about your way. It's my video right here."
Goddamn menace.
This guy is legend.
I was searching for quicker way for algebric twin questions for my entrance exam and found this with "fast" in thumb nail. Yes it was really fast 😂
🤣🤣 yeahhh
Wow, you redeemed yourself by the most aggressive beautiful fast way,
GOOD WORK
Steve why were you triggered on this video? =(
WHEW!!!! My heart is racing 100 mph!!!
: )
This guy is such a beast
You can take x^2 common in the first step. Then you multiply and divide by 2 . Then add and substract 1 in the numerator . And so you have (x+1/x)^2 - 2 in the denominator and 1-1/x^2 in the numerator and and the integral becomes that of arctanx.
And for the second part you have (x-1/x)^2 +2 in the denominator and 1+1/x^2 in the numerator/ use substitution and it becomes a partial fraction and you can integrate to give a logarithmic function .
There is no need to factorise the expression.
LMAO I never thought I'd love a math tutorial vid so much
I never thought I'd love a comment this much too! thank you!
Who is LMAO ?
your videos make math look really fun again...
A living legend 😍
04:11 ~ 04:43 (from transcript)
"and now let's combine all the x squared term namely we have this right here let's just put on the liner this until I got this and then this is another two right here right so we have this Plus this Plus that but that's that sorry this minus this and the plaster a plaster right so in total we have to a minus square root of 2 C and a plus B plus D oh this right here is the coefficient of x squared on the right hand side but once again we don't have x squared on the left hand side that means oh this has to be 0 that's good"
idk why, but its one of my favorite videos on this channel...^^
Johann Bauer thanks!
It could be done simpler: the numerator of the intetrand 1=1/2[(1+x^2) + (1-x^2)], then divide x^2 to both numerators and denominators before introducing new variables u=x-(1/x) for the first part integral and v=x+(1/x) for the second, they will end up with two results: one arctangent function and one logarithmic
We can do it in a simpler method, add subtract x²
Seperate into 2 terms
1/2[(x²+1/x⁴+1)+(1-x²/x⁴+1)]
Get x² out u get
1/2[(1+1/x²)/(x²+1/x²)+(1/x² - 1)/(x²+ 1/x²)]
And reduce denominator to
( x+1/x)² -2
And take the numerator inside d
(Substitute t=numerator and make it dt)
And u get a form of
{dt/t²-a²+ {dt/t²+a²
Then u can apply the respective formula and get the answer
6:32 I love how angrily he says “at all”
In Russia, they usually put square roots in the numerator. The number 1/(2√2)=(√2)/4 is shorter this way
So good bro!!!!! Thank u for solution
You are amazing. Thank you for this and other quality math videos!
Good work, clear and efficient. saved a lot of time for me from tedious hw in which sense you are saving life. ^^
www.quora.com/What-would-be-the-integration-of-frac-1-1-x-4
this one only take you 2min, im saving life now^^
And in 12th grade they give us only 4 marks for this type of long ass integration . Smh
Deep Mondal its very easy
This was so hard! (Even though I was doing this, but at times, I was SO FAR BEHIND that I had to pause the clip just so I could catch up with you!) (BTW, I posted this from Iowa at 5:15 pm Central Time.)
Ah I see. It actually took me a loooooooooong while to practice this integral before I could record.
Happy Thanksgiving Keegan.
Solve sqrt(tanx) with partial fractions! Because it simplifies to 2u^2/(u^4+1) which can be factored
Your work is fine in terms of getting a process that the integral is solvable. I appreciate it but, this problem has a very short and delicate solution with fewer steps of computation.
Thanks sir it helps me a lot
Good idea to calculate this integration
easier by factoring an x^2 from the denominator,and your integral becomes (1/x^2)/(x^2+1/x^2)dx than the denominator becomes either (x+1/x)^2-2 or (x-1/x)^2+2 and then force the derivative of the denominator to the nominator and done
How do you mean “force the derivative of the denominator to the nominator”?
Marcus Åkerman the derivative of (1/x+x) is - 1/x^2 +1 , and since you factored x^2 from the denominator you formed that derivative, and then you add and subtract 1 and you formed the integral in terms of itself and an integral which has the form of f'/f which is lnf
Marcus Åkerman i could send you a pic on social media if you want
@blackpenredpen you are the best
18:02
The happiness of finishing this integral
"If you talk fast enough or work fast enough this isn't that bad AT ALL" ♡
This is the first time I'm frightened by a bprp video
Thanks sir for your valuable work.
Faaast I like it 👍🔥🔥
Write numerator= 1/2((x^2+1)-(x^2-1))
And break the denominator, you will get two standard forms of integrals and you will get your answer in 2 minutes with 2 basic steps
I decided to do this one on my own to see if I could do it and I ended up turning it into 1/2*integral(sqrt(cot(v))*dv) (where tan(v)=u and u=x^2) and I ended up having to watch your video on sqrt(tan(x))
I'm not sure which method is more difficult but I do like how the trig sub worked out (I also ended up using w, q, and r subs)
Bro u r the best. If I were in your school, I'd be in your gang
While the algebra is certainly interesting, this problem would become so much easier through the initial substitution x=t/SQRT{2} because then the denominator becomes t^4+4 which is perfectly factorable into the product of two quadratic polynomials. At the end you only have to backsub.
This was very enjoyable to watch.
What a beastly solution. Geez!
thank you for your answer
哈哈，赞！这个风格不错！数学没必要非是那样死板严肃，有个性带情绪的数学才是人的数学而不是数学家的数学。喜欢你的视频，赞！
xingguang yu 謝謝！ 你也可以看看我朋友peyam的頻道,看他講數學我都也會發笑的
blackpenredpen what is your name?! I've been looking for it. You're bringing back the passion i used to have for math :)
Prof. Steve Chow, M.Ed.
@@douro20 Dude you can't just dox people like that
06:33 HOLY I FELL FROM THE CHAIR
11:19 Did he make the same mistake twice? Either way, thank you for the video, you're making me want to do some integrals again.
Thank you for uploading this video! My teacher told me that I would have to solve this integral on next exam. I'm sure I will get high score.
Well done
I tried to solve this using a x²=tan ø trig sub. Which led to 2 times the integral of a reciprocal nightmare.
Brilliant, thank you!
You could write numerator as1/2 (x2+1) - 1/2(x2-1) and divide both numerator and denomitor hy x2 , easy
you're amazing!!!! i just wanna say thank you
and the answer is ....1/4\/2(-log(x^2-\/2x+1)+log(x^2+\/2x+1) -2atan(1-\/2x) + 2atan(\/2x+1) + C Prove derivating the one
Fantastic video as usual! I have a (probably silly) question. How do you know, in the partial fractions, that the numerator is a first degree polynomial (Ax+B) and not any other degree? Thanks in advance!
Not a silly question at all! Notice that the denominators in the two parts of the partial fractions were quadratics, i.e. of degree two. That means that the numerator is exactly one degree less, so you need a degree one polynomial as the numerator. If the denominator happened to be a third degree polynomial, then you'd need a second degree one in the numerator, so something like Ax^2+Bx+C.
Reflective Ducky Really well explained. Thanks!!!
very nice explanation!
coldmash thank you