Very nice! The other day I went through the trouble of integrating 1/(x¹⁶+1). It was quite tedious Nice use of the arctan sum identity, I didn't think to do that
This is reminds me of what i've suffered from 2 years ago when i was grade 12 doing mathermetic sh** like this lol. And at the time i could 100% understand this. But now after 2 years of not touching math book looking at this again i can partly understand.
Thre's an easier way to integrate 1/(x⁴-x²+1)(=I) I=1/2∫{(x²+1)/(x⁴-x²+1)-(x²-1)/(x⁴-x²+1)}dx 2I=∫(1+1/x²)/{(x-1/x)²+1}-(1-1/x²)/{(x+1/x)²-3}dx Now, it is easy. Let u=x-1/x => du=(1+1/x²)dx v=x+1/x => dv=(1-1/x²)dx The integral very well simplifies to: 2I=∫du/(u²+1)-∫dv/(v²-3) 2I=arctan(u)-1/(2√3)log|v-√3|/|v+√3|
Then you’d have complex logarithms that you’d probably want to convert back into real functions, which is not exactly easy. Regardless, the video demonstrates the process through calculus topics rather than complex analysis.
@@crayyzen1913I did that and got the answer written in a very different form... Conversion to real functions was as nice as all the steps in this video though (it was just as horrendous)
I got lost halfway when you said that A+C or any other term including them is equal to 0 but B+D is equal to 1, I'd like to know how you prove it with certainty, then again maybe I'm stupid or it's something I haven't studied yet
Let me do a physics focused solution: ∫1/(x⁶+1)dx Lets assume x is a great number, this way we can say that x⁶+1~x⁶ So, ∫1/(x⁶+1)dx= ∫1/x⁶dx = ∫d(-1/(5x⁵)) = -1/(5x⁵) + C Since we are assuming x is a great number, the integration constant is negligible, therefore our final solution is ∫dx/(x⁶+1)=-1/(5x⁵)
Easiest math Olympiad question be like:
This is easier than the easiest olympiad question ever, just the matter of doing the calculations correctly, otherwise it’s a straight forward thing
Yeah this kind of problem is more tedious than it is difficult.
나는 많은 계산뿐만 아니라 발상적인 식 변형이 이 문제를 어렵게 한다고 생각했는데, 그렇지 않은가요?
@@quodlibet170for PPL who prep for Olympiads it's like doing basic arithmetic.
Your mom: why does my son not have any friends?
While we’re all in holiday break bro is grinding that calculus
Top 3 math nerds I've seen:
1. Hadi Ridawi
2. Balckpenredpen
3. 3Blue1Brown
I really like the video : everything is nice : the music, the rythm of algebric transformations and the method is nice !
fuck you, wtf do you mean it was nice there was no explanation for anything
Teacher: The test is only problem
The problem:
they watch:😁
they learn:☠
The nostalgic music damn 🗿
FR
Very nice! The other day I went through the trouble of integrating 1/(x¹⁶+1). It was quite tedious
Nice use of the arctan sum identity, I didn't think to do that
X¹⁶....are you alright bro?
@@DKAIN_404 nope!
@@gabrielpartin3474what grade math is this... im already losing it at 10th...
this uses calc 2 concepts, so it just depends on when you take calculus, it could be in 9th-12th grade or in college
This is reminds me of what i've suffered from 2 years ago when i was grade 12 doing mathermetic sh** like this lol. And at the time i could 100% understand this. But now after 2 years of not touching math book looking at this again i can partly understand.
Thank God I understand this
Ok then explain it
@@Ivan-official_a YEAH BABY
If this was one of my question in the exam, imma set the whole campus lit
3blue1brown: lets take this fuction and using some simple caculus, we get:
This was great. Fascinating ❤
Thre's an easier way to integrate 1/(x⁴-x²+1)(=I)
I=1/2∫{(x²+1)/(x⁴-x²+1)-(x²-1)/(x⁴-x²+1)}dx
2I=∫(1+1/x²)/{(x-1/x)²+1}-(1-1/x²)/{(x+1/x)²-3}dx
Now, it is easy.
Let u=x-1/x => du=(1+1/x²)dx
v=x+1/x => dv=(1-1/x²)dx
The integral very well simplifies to:
2I=∫du/(u²+1)-∫dv/(v²-3)
2I=arctan(u)-1/(2√3)log|v-√3|/|v+√3|
If this is what algebra 2 looks like, im done for.
It's calculus
In the US curriculum, this will be in Caculus 2-3 (college LV).... algebra 2 the end of practical math... anything beyond d becomes advance
very nice! transformation is so cold!
I would honestly use trionometric substitution, more straight-forward
What trigonometric substitution would you use? I do not believe there is one that makes the problem easier rather than more difficult
@@hadirihawi Can x^6 = tan^2(u) work?
nice, can you do 1/(1+ x^5) next ? :)
Yes, this will be the next video.
1/1+x^5 is.. monstrous
Create the video: "How to approximate everything like 1+1 or else"
Why can't we use the ln() function?
We do, here it's "log"
Only works for linear functions, you'll get a 6x^5 on the numerator
умоляю, делайте такие видео снова!
Can you factorise x^6 +1 and then use partial fraction decomposition?
Did you watch the video? That’s how it’s done.
@@hadirihawi I mean with the complex roots
Then you’d have complex logarithms that you’d probably want to convert back into real functions, which is not exactly easy. Regardless, the video demonstrates the process through calculus topics rather than complex analysis.
@@crayyzen1913I did that and got the answer written in a very different form... Conversion to real functions was as nice as all the steps in this video though (it was just as horrendous)
Hey bro, what exactly do u doing in your "approximation" videos? How can u 1+1 make that hard, how it possible? =) @@hadirihawi
I would like to know which program you used to create a video like this?
He used Manim, which is a python library for creating mathematical animations
Manim 👍
@@mzm_de which founded by 3blue1brown
Longest Calculus
Could you also solve this by substituting x^3 for tan?
No, that makes it worse.
Its been a year i last solved it
Really was a good problem
It was in our highschool finals
the formula for the definite integral from 0 to infinity of 1/(1+x^n) = (pi/n) * csc(pi/n),
This gotta be related to Euler’s reflection formula right?
@@NintendoGamer789 yes. But I derived it from the betafunction integral
Yes, the standard way to obtain this result is by the residue theorem; this result will be demonstrated in a future video.
Hello, what about int cosx/(x^5+1)? More videos about cosx/(1+x^2) and not cosx/(1+x^5)
How did i end up with x- xlnx/6 😭
Try again.
Chin tapak dam dam😅
Hai koi esa
(x⁵+1)½ ??? I don't English
What you use to make math edits?
Как анимацию сделали ?
Hello. Where are you from. You look like Arabic. Middle East region, right?
Great vid 🎉
Funfact:calculator and google cannot find the solutions
My friend did this and it took him 2 days 😂
Just use the algebraic twin type on the first one could’ve been done in less than 6 steps
No residues???
❤music 2016 again
Buenísimo
Lástima la POLUCIÓN SONORA
Is it possible to integrate 1/x⁷-1 dx
Yes.
I got lost halfway when you said that A+C or any other term including them is equal to 0 but B+D is equal to 1, I'd like to know how you prove it with certainty, then again maybe I'm stupid or it's something I haven't studied yet
Look at the previous step and see if you can deduce this on your own.
@zoja see both side of equation and compare the quantities
This topic is called "Partial fractions" in "Integration" topic
Calc 2 any% speedrun glitchless
complex analysis gonna destroy this (the answer is π/(6sin(π/6))=π/3)
coreection: I thought the integral is from 0 to inf.
😂😂
You can just integral 1/(x^3)^2+1^2 Dx it
No, you can’t.
mind elaborating?@@hadirihawi
@@ye9501 You need to do some sort of substitution, and the problem will become more difficult. You can’t just integrate that as arctan(x^3).
@@hadirihawiwhy?
To put it simply: because of chain rule.
❤❤❤❤❤
Nice 😊
Complex integration methods not be like:
They are like:
Weeeeeeeeee!!!! That's all, this is the answer!
can you do more hard integration videos
Respect
why not ln(x^6 + 1)
you did not just say that
@@jyotishandvlogs801 isnt that the antiderivarive?
@@MartijnMolsNo, by chain rule you have to multiply by the derivative of x^6 + 1, which would give you 6x^5/x^6+1
/ 1/x^6+1=💀
Thanks
No problem!
My dumbass thought this was [ln(x^6+1)]/42
Hi im a sub 😊
Hi!
= x/x⁶+1 🙂
Within 2 mins😂
Hello
Wtf i just watching
why did u use log instead of ln?
Log denotes the natural logarithm in mathematics.
@@hadirihawi no, log(x) is equivalent to log10(x) 😐? and lnx is loge(x). it's different
@@hanvu3343yes you are right
@@hanvu3343 some math papers denote lnx as straight up logx. in chemistry however, log means base 10. its just how its used in question papers.
@@hanvu3343i think it’s fairly clear, what is meant here, so there’s no reason to cry about it
what
No way
Let me do a physics focused solution:
∫1/(x⁶+1)dx
Lets assume x is a great number, this way we can say that
x⁶+1~x⁶
So,
∫1/(x⁶+1)dx= ∫1/x⁶dx = ∫d(-1/(5x⁵))
= -1/(5x⁵) + C
Since we are assuming x is a great number, the integration constant is negligible, therefore our final solution is
∫dx/(x⁶+1)=-1/(5x⁵)
No music? Snore
Whut
To be honest, it’ll be easier to understand if you remove those useless visual effects.