Integral of 1/(x^6+1) without partial fractions!
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- Опубліковано 26 гру 2018
- 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!!
Integral of 1/(x^6+1) via partial fraction & Gaussian Elimiation, • Gaussian Elimination M...
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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: ua-cam.com/video/eugIOsqaI8Q/v-deo.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
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!!!🥂
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
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
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
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!
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.
This guy's enthusiasm about solving integrals is amazing. I love watching this channel
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.
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.
I wish I had you as my recitation teacher. You're an effective teacher!
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
Excellent video. Everything makes sense. Thanks for posting!
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:):):)
Nice approach to question..Love from INDIA 🇮🇳 😊
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!
Dude, u're great, truly love your videos!
Your technics are amazing
Well done. I enjoyed your enthusiasm.
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 ua-cam.com/video/GPvN5UWJlmE/v-deo.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 better now than before. Congrats!
There is another formula for integral of 1/x^2-a^2 which is 1/2a X ln |(x-a)/(x+a)|
Lots of people are not familiar with inverse hyperbolic trigs. Thus we can use partial fractions to compute that 2nd integral
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.
in reality during practice i just differentiatated the 4 options and checked if any of them match the integral, instead of actually integrating
"And let me curse this integral now" you are so funny I like it 😂
it reminded me of the laplace transform when you were adding zeros in the numerators and denominators
Love that hyperbolic trig identity, nothing more satisfying than integration.
You know things will get funny when there are not only the black and the red pen 😂😂😂
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 !
Amazing!
4:58 never thought of that, genius
Me and my friends love your channel
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.
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: ua-cam.com/video/eugIOsqaI8Q/v-deo.html
Thank you
This is Pretty cool ♥
Brilliant !!!
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 UA-cam videos : )
for |x|1 you can use tanh^{-1}(-1/x)
You really have a good Algebra
So good😄
Good explain brother
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?
Great job. I just started to rewrite the integral with pen and paper :). Being more interested in the steps for solving it. I didn't quite get the final answer, totally, but that's a good way for me to start and study more :). Thanks for your effort.
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.
Such a big complicated answer for a seemingly simple integral. Just imagine if the +1 wasnt there it would be so much easier
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) , ∆
u so clever thanks for the solution
God damnit i remeber doing this integral and really getting to know partial fractions afterwards, the algebra way is soooo much better lmao :D
one always likes to be on the top.! /// your videos are really awesome ^_^
Very nice sir thnk u
Felicitacioes muy buena tu integral.
Subscribed!
Одлично, ја сам одушевљен!
Very very Fantastic
Love it
5:02 turn on the subtitles
Me with my middle school maths knowledge: how the heck did you get tan in an equation like that.
And there's ur 10 min gone with 1hr for math section in the exam having 30 questions I think 😆
it's beautiful !
Thank you sir..
i love that mike he holds in his hand
Cool answer, but is there any way to evaluate the definite integral from 0 to infinity on board?:) thx
At 10:43. Why dont you use the identity integral
1/(x^2-a^2). = (1/2a) log((x-a)/(x+a))
Where is the video for the inverse hyperbolic tangent formula you used? Have you made it?
ua-cam.com/video/GPvN5UWJlmE/v-deo.html
@@GhostyOcean thx
blackpenredpen always has interesting math problems that require a lot of thinking.
You are meant to solve this in 59 sec
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.
Twice is better than once, isn’t it?
Oon Han it is!!
Is this a k-pop reference?
Hey, you could have wrote x^6 as x^3 the whole square and then used the direct formula of 1 over x square + a square where x is x^3 and a=√1 right?
multiply and divide by x²+1
I like you channel bro !
It is just awesome and make more videos of jee iit maths.
sorry for my ignorance, but why isn’t he using integration by parts? is it because it can’t be applied or just to show this method?
Love u man
Integrate
1/1+x^6+x^3
If u can. Lets see what your made of
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
Why did you not use the formula for Integration of 1/x^2+a^2 = 1/a arctan x/a + C?
Love the matrix one more XD but could you make a proof that the two answers are the same? Or just off by a constant?
Great video. Just curious, do you teach at a university?
Hey how about taking x^3=tan< X>??
How I learned to resolve integrals of the form 1/(z²-a²) is 1/2A(ln((z-a)/(z+a)). I wonder if it's related to what was written around 11.16
how did you convert x^6 + 1 into the two multiplikants? This eludes me but it is obviously correct.
1/(x^3)^2+1^2
Use the property
Saved your time
In what way is this a Jee integral? My school exam had a similar question to this.
Please do an integral of log(cosx) from [0-(π/2)]
Most useless things to do in life
How did multiplying by 1/2 eliminate the need for writing -1 around 4.58. I don't get it.
are you sure it is tanh^-1 ?
(x+ 1/x)^2 minus (sqroot of 3)^2
can you show me a video of how a tanh inverse is the answer?
Please and thank you.
Very good :)
Can u explain the integration of root over of tan^-1 with limits -infiniti to +Infiniti ?
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.
Could this be solved by taking x^2=u then do partial fraction?
2:01 hey how does x2 -1 cancels?
Wao nice really great
If i remember well i saw that he caluculated this integral also by partial fraction decomposition