Integral of 1/(x^6+1) without partial fractions!

[@blackpenredpen]

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

[@gazalisameer4173]

Can someone help me with integral 1/sqrt(a + bsinx)

[@axelcastillo7432]

blackpenredpen But, isn’t coth^-1 equal to tanh?

[@fmcore]

@@axelcastillo7432 No. it is not reciprocal. It's an inverse function.

[@rushikesh3363]

Your note is totally wrong. .. Use their standerd FORMULA

[@enriqueiniguez6893]

GAZALI SAMEER 2(a+bsin(x))^1/2+C

[@arjavgarg5801]

Put more JEE in the title = more views

[@integrationbee2511]

You can see my channle for JEE integrals. I have already solve this integral few months ago. See: kzbin.info/www/bejne/m6bKeoKppsZ8brM

[@arjavgarg5801]

Cauchy Riemann I am a JEE aspirant too BTW

[@0ArshKhan0]

@@arjavgarg5801 Oh, all the best for jee, I'm a first year student at bits.

[@arjavgarg5801]

Cauchy Riemann did you sit for it? What rank?

[@0ArshKhan0]

@@arjavgarg5801 A useless 10k+ doesn't get you anything in IITs. Anyway, let's not dig the past :-P

[@bapolino733]

Sleeping at 3 am ? No Math is more important!

[@DiegoMathemagician]

Yay!

[@diegofcm6201]

Bapolino 1:30 for me, lol

[@ricardomahfoud]

Ikr I have been up all night, 6am watching this right now

[@WardForFun]

@@ricardomahfoud DUDE ME TOO

[@sowmyag5142]

Yes!!!🥂

[@rishavsinha3376]

It's a custom for physics students to binge watch videos on difficult integrals.. You never know when it will come in handy.

[@sheetalprakash9827]

I don’t know anything about Calculus but I just like staring at all those letters and numbers.

[@DragonKidPlaysMC]

Microwave Burrito you should check out the essence of calculus by 3blue1brown

[@sheetalprakash9827]

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.

[@josephkitchen3059]

It’s how I fell in love with math.. became an engineer. Just a curiosity.

[@sheetalprakash9827]

Artorias Thugz Oh cool. Hope you have a great life.

[@shilpimitra5342]

4 years back i was just like u

[@martinmetodiev869]

This guy's enthusiasm about solving integrals is amazing. I love watching this channel

[@ssdd9911]

i DARE u 2 do the integral of 1/(x^8+1)

[@blackpenredpen]

If you make a tweet asking me to integrate 1/(x^8+1) and the tweet gets 2019 likes, then I will. : )

[@cavver3523]

That's a great idea lol

[@Ferolii]

@@blackpenredpen You will be asked soon to the 1/(x^10+1) integral hahaha

[@visheshmangla2650]

Can be done by contour integration or just break it to a^2- b^2 using complex numbers.

[@Jacob-uy8ox]

This would make me cry of happiness, like if you like super insane af integrals!

[@arjavgarg5801]

Yeah, I would rather just differentiate the options

[@deepcvs]

You would take a lifetime

[@satyam2857]

That's my way..

[@satyam2857]

Anudeep CVS it takes 10 minutes.. So if you are done with all questions.. U can do that

[@mdghufranalam7369]

I know but you will loss time if a small mistake happen. Differentiate the same answer and get question you cannot simplify

[@arjavgarg5801]

Not in mains

[@angelmendez-rivera351]

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).

[@rayandy2460]

I wonder is there a recursive formula to derive the above integral.

[@jimspelman8538]

@@rayandy2460 I was wondering the same thing, but I don't think there is.

[@obiwan8972]

Now this is a standard JEE integral:)

[@blackpenredpen]

Obi Wan Yup

[@leif1075]

@@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.

[@bigbrain296]

@@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.

[@leif1075]

@@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.

[@wqltr1822]

@@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.

[@oledakaajel]

First blue pen and now green pen? This is getting out of hand.

[@cavver3523]

BlackpenRedpenBluepenGreenpen. It's going to be a Rainbowpen

[@chiragraju821]

NOW THERE ARE TWO OF THEM

[@avaneshprasan6492]

from the screen to the ring to the pen im the king

[@sreenandhan2544]

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!

[@dabulls1g]

Feels awesome your integration tables know about hyperbolic tangent, not you lol.

[@sreenandhan2544]

But does it say what hyperbolic tangent really is? :)

[@abdiismail4546]

I don't think so. The derivative of ln {(u-a)(u+a) } = 2u/(u^2 -a^2)

[@tomatrix7525]

This is the approach used on wolfram, yes. Nice observation

[@superj9220]

With all these similar ones i wanna see a 1/(x^n +1) integral generalization lol.

[@TheYou1483]

Don't know about the indefinite but it's actually generalized for definite integral with limits ranging from 0 to infinity

[@boringtofu4433]

I remember doing a reduction formula for this its pretty easy to prove

[@arvindupadhye6172]

For definite integral with limits 0 to infinity, it becomes the beta function

[@soheilshirmohamadi3449]

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☺

[@blackpenredpen]

Ahhhh, I forgot to check the min. of (x+1/x)/sqrt(3) You are right, I should have used arcoth, thank you!!!

[@blackpenredpen]

For anyone who's interested, check out the derivative of arctanh vs. arcoth kzbin.info/www/bejne/fYHZf2iLjK-fo6c

[@soheilshirmohamadi3449]

@@blackpenredpen oh no problem, in fact thank you for these fabulous math videos:-)👌👍

[@officielsalah5838]

Your note is not complet you forget u' in the top

[@kartikkalia01]

I liked this comment to just look smart.

[@RaviKant-vn5wb]

Nice approach to question..Love from INDIA 🇮🇳 😊

[@shreyasgavhalkar57]

There is another formula for integral of 1/x^2-a^2 which is 1/2a X ln |(x-a)/(x+a)|

[@nicolesmith2782]

I love your videos! I decided to take on calculus 2 this semester and you motivate me to continue with math :))

[@blackpenredpen]

Awww thank you! I am very happy to hear it! Best of luck and enjoy calc 2!

[@animeshpradhan5683]

Please do the integral 1/(x^7+1)

[@123pok456ey]

Or find the general formula of indefinite integral of 1/(1+x^n)

[@animeshpradhan5683]

I think it will b a reduction integral

[@123pok456ey]

I've done in in the complex world.

[@animeshpradhan5683]

@@123pok456ey nth root of unity.. ?? De moivres

[@marks9618]

wc k Residues?

[@peerdox2275]

in reality during practice i just differentiatated the 4 options and checked if any of them match the integral, instead of actually integrating

[@Grassmpl]

Lots of people are not familiar with inverse hyperbolic trigs. Thus we can use partial fractions to compute that 2nd integral

[@MrKhan-dw9vh]

"And let me curse this integral now" you are so funny I like it 😂

[@metalslug97]

I wish I had you as my recitation teacher. You're an effective teacher!

[@ahb5819]

You are meant to solve this in 59 sec

[@jadsonalves7590]

You know things will get funny when there are not only the black and the red pen 😂😂😂

[@javierfromvenezuela]

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?

[@ernestschoenmakers8181]

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.

[@emmanuelokafor2598]

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.

[@Peter_1986]

blackpenredpen always has interesting math problems that require a lot of thinking.

[@dylanf2980]

I got an eel i buried its guts sprouted a tree now you got coconuts.

[@soulofraven816]

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 !

[@themanagement69]

Love that hyperbolic trig identity, nothing more satisfying than integration.

[@maehmaehmaeh9560]

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

[@ernestschoenmakers8181]

You're wrong cause you change the integral into a completely different one which isn't the same anymore.

[@prahladverma7979]

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:):):)

[@ronjamac]

Wow. This takes me back to my school days in the 1960s. Just wondering now where I could find a real live practical application.

[@blackpenredpen]

ronjamac Making KZbin videos : )

[@pratyushsharma6655]

And there's ur 10 min gone with 1hr for math section in the exam having 30 questions I think 😆

[@yaleng4597]

4:58 never thought of that, genius

[@JohnSmith-iu3fc]

You are better now than before. Congrats!

[@ghotifish1838]

Me with my middle school maths knowledge: how the heck did you get tan in an equation like that.

[@chessandmathguy]

Excellent video. Everything makes sense. Thanks for posting!

[@shivimish9962]

Where is the video for the inverse hyperbolic tangent formula you used? Have you made it?

[@GhostyOcean]

kzbin.info/www/bejne/fYHZf2iLjK-fo6c

[@shivimish9962]

@@GhostyOcean thx

[@pranjaldas1762]

Your technics are amazing

[@AbouTaim-Lille]

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) , ∆

[@chahbiachraf5703]

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 ....

[@Grassmpl]

How? This isn't a definite integral. There are no bounds.

[@OonHan]

Twice is better than once, isn’t it?

[@blackpenredpen]

Oon Han it is!!

[@hiran1724]

Is this a k-pop reference?

[@hamzaalsamraee3054]

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.

[@Infinite_Precision]

Dude, u're great, truly love your videos!

[@stephenphelps920]

5:02 turn on the subtitles

[@jianyuchen8879]

You really have a good Algebra

[@ashishgiri2373]

I dont know is it right or not but formula wil work easily 1/x2+a2would be tan ^-1(x/a)1/a will it be right

[@hashblack11]

Exactly i thought of the same!

[@hashblack11]

I thought the answer was supposed to be tan^-1 x^3 +C

[@ernestschoenmakers8181]

@@hashblack11 Incorrect.

[@r.t.mishra7786]

This is one of the easiest problem of the toughest exam of the world iit jee

[@ojasthengadi9681]

Lol kid this not at all easiest this quite a good one and almost falls into tough category bcoz the amount of time it takes to solve it

[@marcovillalobos5177]

Me and my friends love your channel

[@rocketboyjv5474]

Such a big complicated answer for a seemingly simple integral. Just imagine if the +1 wasnt there it would be so much easier

[@AhmedHan]

First 1/(1+x^2), then 1/(1+x^4), and now 1/(1+x^6). What is next, 1/(1+x^8)? For the sake of goodness, jus solve 1/(1+x^n) once and for all!

[@ernestschoenmakers8181]

It's not that simple the higher the nth power the harder it becomes but in general it can be solved by using complex numbers.

[@sudhanvab]

At 10:43. Why dont you use the identity integral 1/(x^2-a^2). = (1/2a) log((x-a)/(x+a))

[@dylanjones4915]

"One always likes to be on the top" -blackpenredpen

[@holyshit922]

for |x|1 you can use tanh^{-1}(-1/x)

[@qu2k458]

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.

[@carolynrigheimer1574]

Well done. I enjoyed your enthusiasm.

[@FernandoRodriguez-et7qj]

Your shirt makes you look like a nurse

[@sowhanQ]

pretty sure this appears in my exams once

[@rene_mpo9643]

Can you do the integral from e to pi x^y=1 solve for the y

[@angelmendez-rivera351]

What? Your question does not make sense with the phrasing.

[@0ArshKhan0]

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

[@binitkumarsingh8296]

U are a good technical teacher.u should start teaching for jee preparation

[@amoghbharadwaj382]

No way. Don't drag him into that mess. This channel is for math enthusiasts not math exam enthusiasts

[@richardbaek976]

I wish i could understand this beautiful language

[@XanderGouws]

idk if ur memeing, but you'll get there eventually! There are plenty of resources to learn calculus online. I personally reccomend "Professor Leonard" (KZbin) and "Paul's online notes" (website).

[@anand.suralkar]

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

[@kanishksharma1716]

@@anand.suralkar except those who take arts and commerce without maths.

[@Goku17yen]

God damnit i remeber doing this integral and really getting to know partial fractions afterwards, the algebra way is soooo much better lmao :D

[@omkarsinghchauhan3053]

i love that mike he holds in his hand

[@AS-hm4sh]

This is Pretty cool ♥

[@shravankumardilip7146]

Please do an integral of log(cosx) from [0-(π/2)]

[@the_number_one]

Most useless things to do in life

[@andres.robles6]

Más integrales así 💙

[@emmanuelokafor2598]

How did multiplying by 1/2 eliminate the need for writing -1 around 4.58. I don't get it.

[@GreenKookie56]

With that being said, the intergral of 1/1+x⁶ is 50 times less trivial than 1/1+x⁵.

[@SuperKnowledgeSponge]

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.

[@golddddus]

Одлично, ја сам одушевљен!

[@skippycavanaugh3148]

In what way is this a Jee integral? My school exam had a similar question to this.

[@harshaddeshmukh3753]

Advance paper is held each year you have a lot of content.

[@blackpenredpen]

Link please?

[@harshaddeshmukh3753]

@@blackpenredpen you just type on Google "IIT JEE advanced paper 20xx and you can download it.

[@adityashankar5267]

2:01 hey how does x2 -1 cancels?

[@snnwstt]

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.

[@chunfaimok767]

1:11 1 always likes to be ON THE TOP www. I love you, man.

[@blackpenredpen]

: )

[@Jacob-uy8ox]

Omg, what a beautiful way to end up the year

[@sowmyag5142]

Yay!!

[@afafsalem739]

Great job, plz we would like more of this kind of integrations , thank you very much.

[@integrationbee2511]

You can see my vedies. Such: kzbin.info/www/bejne/m6bKeoKppsZ8brM

[@greatestever6983]

it reminded me of the laplace transform when you were adding zeros in the numerators and denominators

[@mamtaprajapati2710]

Can you integrate sin(x^x) please? Or sin(x^x)/(x^x)....

[@SonGoku-rx1fo]

It would be sin (x^x) Sin(x^x)= x^x sin(x^x)[ 1 + log x ] I guess I'm right Sin(x^x)= sin (x^x) [ 1 + log x ] I guess I'm right

[@SonGoku-rx1fo]

It's not integration btw it's differentiation , ok ?

[@mamtaprajapati2710]

@@SonGoku-rx1fo But still thanks for that....

[@anand.suralkar]

@@mamtaprajapati2710 lol

[@amoghbharadwaj382]

@Simon Slade Bro no offense but can you not read? It clearly says "integrate" in the comment not "differentiate"

[@PranavvSalunkhe]

It took him 13 minutes to solve this questions whereas in reality students have to solve it in less than 2 minutes ....

[@blackpenredpen]

Yea.

[@axelcastillo7432]

Ctrl+c Ctrl+v Takes less than 30 seconds, it won’t look fishy in the test, i promise lol

[@pramodm3540]

Because he has to explain it. I know 2min is challenging.

[@paulfaigl8329]

how did you convert x^6 + 1 into the two multiplikants? This eludes me but it is obviously correct.

[@shobitsagar1211]

multiply and divide by x²+1

[@TheDoubleEdgeSword]

2nd line....how x^2/(x^3)^2+1 came from?....and where is x^+1 of the first line?

[@Patapom3]

Amazing!

[@Aruthicon]

I like how you wrote the inverse hyperbolic tangent backwards.

[@PeterNjeim]

I like how our standard notation has us write the inverse hyperbolic tangent backwards.

[@theomolfess7272]

No it’s only because you speak backwards in English. For example, in French we say tangente hyperbolique ;)

[@AnuragKumar-io2sb]

So good😄

[@yuvrajpreetsingh591]

Why did you not use the formula for Integration of 1/x^2+a^2 = 1/a arctan x/a + C?

[@YorangeJuice]

4:39 what was that sound?

[@emmanuelokafor2598]

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

[@Fatihuuuu]

bu adam bizim ayt matematik sınavına girse 10 net yapardı

[@LaerteBarbalho]

Black pen Red pen, also featuring Blue and Green pens...😄

[@Sriiiiiiiiiiiiiiiiiiiiii123]

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?

[@franciscosoto2276]

Integral of 1/(x^n + 1) next! That would be nuts!

[@ernestschoenmakers8181]

Thuis cannot be solved in a simple wat but it can be done by complex numbers.

[@chestercarbungco2967]

1:02 pls answer me where -x^2 came from? Pls need help. Thanks!

[@radhikasomani4642]

He just created a difference by adding x^2 and subtracting x^-2

[@abdraoufabusoua8004]

This is amazing. Isn’t math beautiful? .

[@kurosakitong]

Why didn't he add +1 at 2nd step ?

[@BigDBrian]

Maybe it'd be good to define a function that's basically the inverse hyperbolic tangent & inverse hyperbolic cotangent together. Since their domain doesn't overlap, it's a simple to define piecewise function.

[@angelmendez-rivera351]

mrBorkD This is easy to do with the complex numbers, I am fairly certain. You can define an analogous real-valued function, but it would not be a very nice function to work with.

[@angelmendez-rivera351]

Actually, it is better to not reference the inverse hyperbolic functions for this at all and instead use partial fraction decomposition. The catch is that [(tanH)^(-1)](z) + sign(z - 1)•iπ/2 = [(cotH)^(-1)](z) is true for all complex z. This is because the first function is Ln[(1 + z)/(1 - z)]/2 while the second is Ln[(z + 1)/(z - 1)]/2, in other words, with the negated input, but the catch is that these identities are only true when working with complex numbers.

[@ThePeterDislikeShow]

Can you use imaginary numbers in your partial fraction?