Integral of 1/(x^6+1) without partial fractions!
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- Опубликовано: 4 окт 2024
- No partial fractions in the usual sense, but we still managed to break down 1/(x^6+1) and got to the forms that we needed in order to integrate, enjoy!!
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Note: I should have used coth^-1 instead since the input 1/sqrt(3)*(x+1/x) is NOT in between of -1 and 1
Can someone help me with integral 1/sqrt(a + bsinx)
blackpenredpen But, isn’t coth^-1 equal to tanh?
@@axelcastillo7432 No. it is not reciprocal. It's an inverse function.
Your note is totally wrong. ..
Use their standerd FORMULA
GAZALI SAMEER 2(a+bsin(x))^1/2+C
Put more JEE in the title = more views
You can see my channle for JEE integrals.
I have already solve this integral few months ago.
See: ruclips.net/video/eugIOsqaI8Q/видео.html
Cauchy Riemann I am a JEE aspirant too BTW
@@arjavgarg5801 Oh, all the best for jee, I'm a first year student at bits.
Cauchy Riemann did you sit for it? What rank?
@@arjavgarg5801 A useless 10k+ doesn't get you anything in IITs.
Anyway, let's not dig the past :-P
It's a custom for physics students to binge watch videos on difficult integrals..
You never know when it will come in handy.
I don’t know anything about Calculus but I just like staring at all those letters and numbers.
Microwave Burrito you should check out the essence of calculus by 3blue1brown
DragonKidPlaysMC
Ok thanks I’ll see. I’m only in grade 8 so I’ll probably start learning it next year but thanks for the video. It might give me a head start next year.
It’s how I fell in love with math.. became an engineer. Just a curiosity.
Artorias Thugz
Oh cool. Hope you have a great life.
4 years back i was just like u
Sleeping at 3 am ? No Math is more important!
Yay!
Bapolino 1:30 for me, lol
Ikr I have been up all night, 6am watching this right now
@@ricardomahfoud DUDE ME TOO
Yes!!!🥂
This guy's enthusiasm about solving integrals is amazing. I love watching this channel
i DARE u 2 do the integral of 1/(x^8+1)
If you make a tweet asking me to integrate 1/(x^8+1) and the tweet gets 2019 likes, then I will.
: )
That's a great idea lol
@@blackpenredpen You will be asked soon to the 1/(x^10+1) integral hahaha
Can be done by contour integration or just break it to a^2- b^2 using complex numbers.
This would make me cry of happiness, like if you like super insane af integrals!
Yeah, I would rather just differentiate the options
You would take a lifetime
That's my way..
Anudeep CVS it takes 10 minutes.. So if you are done with all questions.. U can do that
I know but you will loss time if a small mistake happen. Differentiate the same answer and get question you cannot simplify
Not in mains
Now this is a standard JEE integral:)
Obi Wan
Yup
@@blackpenredpen Who would come up with such a convoluted solution and HOW?? I hope you can PLEASE respond for once, please. It would mean alot.
@@leif1075 well, x+1/x is a standard strategy in a competitive integral environment and the other integrals are standard textbook. The partial fraction in the beginning was also a logical first step. It's all about recognising which technique to be applied where.
@@bigbrain296 i don't see how it could be standard..and if you've never seen it before..would anyone ever think of it? I doubt it.
@@leif1075 I think there are just some people in the right place at the right time, and have a knack for clever puzzles and stuff, and creating problems which have smart solutions. They might not even understand why some rules or tricks are in place, or why they work, but they love using them in smart ways regardless of context.
Those tricks are shared, and after that its one of those things, where after you see it once or twice or thrice, you kind of get the idea. There is not really a world between not knowing the trick and knowing it.
Find the antiderivative of 1/(x^n + 1) with respect to x. Then you will not be asked to do it for 1/(x^8 + 1) and 1/(x^10 + 1).
I wonder is there a recursive formula to derive the above integral.
@@rayandy2460 I was wondering the same thing, but I don't think there is.
First blue pen and now green pen? This is getting out of hand.
BlackpenRedpenBluepenGreenpen. It's going to be a Rainbowpen
NOW THERE ARE TWO OF THEM
Nice approach to question..Love from INDIA 🇮🇳 😊
Integral of ( 1/(u^2 - a^2)) is also equal to {1/2a} {log[ (u-a) /(u+a)]}+c for those who don't know about hyperbolic tangent, like me.
And off course the video was amazing!
Feels awesome your integration tables know about hyperbolic tangent, not you lol.
But does it say what hyperbolic tangent really is? :)
I don't think so. The derivative of ln {(u-a)(u+a) } = 2u/(u^2 -a^2)
This is the approach used on wolfram, yes. Nice observation
With all these similar ones i wanna see a 1/(x^n +1) integral generalization lol.
Don't know about the indefinite but it's actually generalized for definite integral with limits ranging from 0 to infinity
I remember doing a reduction formula for this its pretty easy to prove
For definite integral with limits 0 to infinity, it becomes the beta function
I love your videos! I decided to take on calculus 2 this semester and you motivate me to continue with math :))
Awww thank you! I am very happy to hear it! Best of luck and enjoy calc 2!
I wish I had you as my recitation teacher. You're an effective teacher!
There is another formula for integral of 1/x^2-a^2 which is 1/2a X ln |(x-a)/(x+a)|
in reality during practice i just differentiatated the 4 options and checked if any of them match the integral, instead of actually integrating
4:58 never thought of that, genius
I can't believe I just discovered this channel. This guy is good!! Fucking good I tell you!! You will save my life for the next few years.
I can't understend English very well , but i can understend the integral . So i thought the mathematic is the Universal lenguague :3 🖤 good joob !
Oh bro watch out it can't be arctanh because automatically one over x or (1/x) omits zero which makes the denominator zero and you probably know that arctanh is defined when abs(x)≤1 and this range of x contains zero so that's why arccoth is precisely the one and only choice without having the bounds of integration, hope you'll see this☺
Ahhhh, I forgot to check the min. of (x+1/x)/sqrt(3)
You are right, I should have used arcoth, thank you!!!
For anyone who's interested, check out the derivative of arctanh vs. arcoth ruclips.net/video/GPvN5UWJlmE/видео.html
@@blackpenredpen oh no problem, in fact thank you for these fabulous math videos:-)👌👍
Your note is not complet you forget u' in the top
I liked this comment to just look smart.
You are meant to solve this in 59 sec
Lots of people are not familiar with inverse hyperbolic trigs. Thus we can use partial fractions to compute that 2nd integral
Excellent video. Everything makes sense. Thanks for posting!
Please do the integral 1/(x^7+1)
Or find the general formula of indefinite integral of 1/(1+x^n)
I think it will b a reduction integral
I've done in in the complex world.
@@123pok456ey nth root of unity.. ?? De moivres
wc k Residues?
Your technics are amazing
Love the way you and subtracting values to complete squares! Its very creative and I hope that i might someday be as proficient at it as you are blackpenredpen 🙏🏼. Great Videos.
Love that hyperbolic trig identity, nothing more satisfying than integration.
Wow. This takes me back to my school days in the 1960s.
Just wondering now where I could find a real live practical application.
ronjamac
Making RUclips videos : )
"And let me curse this integral now" you are so funny I like it 😂
i love that mike he holds in his hand
The 1/x^4-x^2+1 part I had solved a few weeks back. I’m glad to see blackpenred use the exact same method I’d figured out then!! Gives me lot of confidence:):):)
You are better now than before. Congrats!
Well done. I enjoyed your enthusiasm.
You know things will get funny when there are not only the black and the red pen 😂😂😂
Just substract x^4 on the top and bottom, then you can factorize the top and cancel the (x^2+1); then you only have to integrate x^2-1 and you get 1/3x^3-x. Its really easy
You're wrong cause you change the integral into a completely different one which isn't the same anymore.
Dude, u're great, truly love your videos!
There are many ways to solve this integral like one can factor 1+x^6 into (1+x^2)(x^2-sqrt(3)x+1)(x^2+sqrt(3)x+1).
Now with partial fraction decomposition this integral can be solved.
You really have a good Algebra
I wish i could understand this beautiful language
idk if ur memeing, but you'll get there eventually!
There are plenty of resources to learn calculus online. I personally reccomend "Professor Leonard" (RUclips) and "Paul's online notes" (website).
You must have been born in India those things are in school in indian maths syllab calculas is the thing that every indian should go through one day in their lifetime
@@anandsuralkar2947 except those who take arts and commerce without maths.
U can resolve it with the theorem of residue (changing the function into f(Z)=1/z^6+1) and it s a holomorphic function ....
How? This isn't a definite integral. There are no bounds.
Me and my friends love your channel
Más integrales así 💙
Great job, plz we would like more of this kind of integrations , thank you very much.
You can see my vedies.
Such: ruclips.net/video/eugIOsqaI8Q/видео.html
We used to study in our first year the integrals of fractions where the numerator is a polynomial of degree strictly lower than the degree of the denominato (otherwise you just devide). And we have to write the denominator as a multiplication of polynomials of degree 1 and 2. Depending on the theorem that every polynimial can be written as multiplication of terms of the form (X+a) or (x²+ax+b) , ∆
Good explain brother
Omg, what a beautiful way to end up the year
Yay!!
pretty sure this appears in my exams once
Thank you
I like how you wrote the inverse hyperbolic tangent backwards.
I like how our standard notation has us write the inverse hyperbolic tangent backwards.
No it’s only because you speak backwards in English. For example, in French we say tangente hyperbolique ;)
u so clever thanks for the solution
One can use a double integral and then the gamma function to get a general form for the definite integral of 1/(x^n+1). The derivation is very elegant and short.
Such a big complicated answer for a seemingly simple integral. Just imagine if the +1 wasnt there it would be so much easier
multiply and divide by x²+1
blackpenredpen always has interesting math problems that require a lot of thinking.
So good😄
God damnit i remeber doing this integral and really getting to know partial fractions afterwards, the algebra way is soooo much better lmao :D
5:02 turn on the subtitles
U are a good technical teacher.u should start teaching for jee preparation
No way. Don't drag him into that mess. This channel is for math enthusiasts not math exam enthusiasts
it reminded me of the laplace transform when you were adding zeros in the numerators and denominators
1:11 1 always likes to be ON THE TOP www. I love you, man.
: )
Amazing!
For the "green" integral better use (1/2a)ln [(u-a)/(u+a)]
Please do an integral of log(cosx) from [0-(π/2)]
Most useless things to do in life
Its is in our maths book in high school
It is vert easy if you know the formula
1/a Tan^-1 x / a + C right?
@@hashblack11 Wrong.
Felicitacioes muy buena tu integral.
With that being said, the intergral of 1/1+x⁶ is 50 times less trivial than 1/1+x⁵.
"One always likes to be on the top" -blackpenredpen
Twice is better than once, isn’t it?
Oon Han it is!!
Is this a k-pop reference?
Одлично, ја сам одушевљен!
Me with my middle school maths knowledge: how the heck did you get tan in an equation like that.
for |x|1 you can use tanh^{-1}(-1/x)
Very very Fantastic
one always likes to be on the top.! /// your videos are really awesome ^_^
This is Pretty cool ♥
If i remember well i saw that he caluculated this integral also by partial fraction decomposition
Black pen Red pen, also featuring Blue and Green pens...😄
Your shirt makes you look like a nurse
I'm self-teaching myself Calculus by going through a Calculus textbook and this channel motivates me to keep going
Brilliant !!!
Nice sir love from india
Your accent is sweet!
Love u man
Integrate
1/1+x^6+x^3
If u can. Lets see what your made of
Subscribed!
I feel that integration by parts could work.
And there's ur 10 min gone with 1hr for math section in the exam having 30 questions I think 😆
I think this is really about factors of red and black pens!
Given that a polynomial of (real coefficients) can be a PRODUCT of polynomial of degree 2 and degree 1, call them the first factors ( a little bit as decomposing an integer into its prime factors);
given that someone can find Pn(x) = Product of the said "first factors", Qi(x) (with the degree of each Qi is at most 2);
given that someone can find the coefficients in the expression 1/Pn = (Ax+B)/Q2i + C/Q1i
(if you prefer: when Qi of a degree 2, is involved in the denominator, then the numerator has a linear term (2 constants) while for Qi being a linear term in the denominator, then the numerator has only a constant);
then, the original integral is reduced to a SUM of integrals of the kind (ax+b)/(x2+cx+d) and of the kind a/(x+b) for any integral degree n ( >=3) of the initial polynomial expression.
Sure, you have to deal with the discontinuities when the limits' range spans over zero(s) of Pn.
Mantap betul👍👍
The second one is the inverse hyperbolic tangent? Oh! WOW!
I like you channel bro !
Same quest was asked in my mcq test
1/(x^3)^2+1^2
Use the property
Saved your time
it's beautiful !
Can you do the integral from e to pi x^y=1 solve for the y
What? Your question does not make sense with the phrasing.
integral from π to e x^y dx = 1
Hence, (e^y-π^y)/(y+1)=1.
Or e^y - π^y = y+1 which can be solved by a numerical method like Newton-Raphson method...
Solution is close to y= -0.934
Sooo long but looks easy to understand
Where is the video for the inverse hyperbolic tangent formula you used? Have you made it?
ruclips.net/video/GPvN5UWJlmE/видео.html
@@GhostyOcean thx
integration if 1+x^5-z^2/x^6
Advance paper is held each year you have a lot of content.
Link please?
@@blackpenredpen you just type on Google
"IIT JEE advanced paper 20xx and you can download it.