integral of 1/(x^2+1) but you didn't learn it this way in calculus 2

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When you want to use complex numbers to integrate 1/(x^2+1)! We didn't use partial fraction decomposition with complex numbers to integrate 1/(x^2+1) in calculus 2. We usually use a trigonometric substitution to integrate 1/(1+x^2), i.e. the tangent substitution. But here the key is to simplify the complex logarithms and get their principal branch. Be sure to look at the polar form of the complex numbers. Then we can integrate 1/(1+x^2) and get arctan(x), just as expected! This is definitely a great challenge question for any calculus student!
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[@blackpenredpen]

Learn more calculus by visiting Brilliant 👉 brilliant.org/blackpenredpen/ (20% off with this link!)

[@diaa5355]

Goodbye head of math 🇨🇳🇨🇳🇨🇳🇨🇳

[@ImranKhan-tu7ix]

In (x+1 )^2 intigration why we use u = x+1 ans is (x+1)^3/3 = x^3/3 + x^2 + x +1/3 Why not use u = x or if we dont use u why answer different Ans is x^3/3 + x^2 + x Difference is 1/3 one have 1/3 other Don't have 1/3

[@felipelopes3171]

Also, if you did not use the principal branch of the complex log, the only difference would be a shift of the integration constant, so using other branches also works!

[@ItsPungpond98]

"I know I know, I know you know, and you know I know." The greatest phrase spoken by this man.

[@JUSTREGULARSCREAMINGAAHH]

Thanos endgame ahh phrase

[@fanamatakecick97]

This was mind blowing. Imagine if this was a bonus question on a test before you get to the inverse trig functions

[@fix5072]

Good lucking knowing that identity without having studied inverse trig functions first

[@fanamatakecick97]

@@fix5072 It’s not knowing the identity if it asks you to “integrate 1/(x^2 + 1)” as the test question

[@fix5072]

@@fanamatakecick97 yeah that'd be fair

[@edgarb.6187]

It would not have worked in my class because they taught us trig sub before partial fractions.

[@krishna2803]

all my friends thought i was doing it wrong by using complex numbers to solve integrals albeit getting the correct results now I'll show this video to them if it ever happens again! thanks bprp!

[@himanshu6002]

They just mad they can't harness the power of complex world 😎

[@sniperwolf50]

The reason you can get away with this method is that the complex integral of 1/(z² + 1) is path independent, which is not generally true for any complex function f(z). Path independent integrals can be computed as if they were integrals in the reals.

[@krishna2803]

@@sniperwolf50 woah thats cool! i think i saw something like this in a Michael Penn video in which he talked about complex integrals and the fact that the result changes when the path is changed but it didn't ring any bell here. thanks! (1) here, Steve mentioned that 'x' is real, so i kinda thought that this is just a 'real' integral with some goofy tricks, in the complex integral case, we would be integrating over a complex variable (idk if thats how to say it) does that make any difference or it's still the same? (2) also, more generally, if we can reduce an expression involving complex numbers to an expression having only reals, will this still hold? like (x+i)(x-i) can be reduced to (1+x²) which is real for all real x if (2) is true, then it would mean that this should work for all real integrals as we are just creating a complex expression from a real one, so it can always be reduced a real expression

[@aav56]

Here's a simple explanation for why tan-1(x) + tan-1(1/x) = pi/2: Imagine a right triangle, "a" units long and "b" units tall. Its angles will be 90°, tan-1(a/b), and tan-1(b/a). As with any triangle, these must add to 180°. 90° + tan-1(a/b) + tan-1(b/a) = 180° tan-1(a/b) + tan-1(b/a) = 90° (= pi/2) Now sub in x = a/b to derive the identity: tan-1(x) + tan-1(1/x) = pi/2

[@TheCrunchyGum]

Nice

[@moviesadda24x79]

Amazing explanation

[@rafaelstv]

Thanks.

[@YoutubeModeratorsSuckMyBalls]

No, it is longer than in video, there is simpler explanation. Tan(theta) = x, cot(theta) =1/tan(theta)=1/x, but cot(theta)=tan(pi/2-theta)=1/x. So Theta= arctan(x)=pi/2-arctan(1/x). So arctan(1/x) + arctan(x) =pi/2

[@pedroribeiro1536]

Beautiful, thanks a lot for the proof

[@zelda12346]

Love it. It clearly demonstrates that we classify/identify function based on how they relate to each other and not necessarily their explicit details. Like how sin and cos are just the same function with a phase shift.

[@DavesMathVideos]

Hahah, today I was doing a lecture about Partial Fractions and one of the students actually tried to do this. I said, "Usually I would tell you don't memorize, but in this case it's faster just to know that integral if 1/(x^2+1) = arctan(c).

[@pi_xi]

I am pretty sure that the integral is arctan(x) + c and not arctan(c).

[@cillipill]

​@@pi_xiyessss

[@kornelviktor6985]

​@@pi_xi bruhh

[@shreyjain3197]

this is definitely a more intuitive approach as compared to substituting x = tan theta

[@givrally7634]

Fun fact, that also explains why the Maclaurin series for the arctangent only has a radius of convergence of 1. If you come back to 1/2i (ln(x-i)-ln(x+i)), you get 1/2i ln((x-i)/(x+i)), which is undefined when x = i (because you get ln(0)) and x=-i (because you get smth divided by 0). So the series centered around 0 will only converge when |x| < 1... because it only converges when |z|

[@gubgubbubbub]

This is quite helpful in computer algebra systems, because integration of rational functions can be reduced to the case of linear, rather than quadratic denominators

[@tortillajoe9942]

I’ve done this very integral the same way before! One might consider the particular solution satisfying y(0)=0 and you get tan⁻¹(x) and (1/2i)ln((x-i)/(x+i))-π/2

[@Batshitcrazy6942]

Bro me too, bro me too.

[@militantpacifist4087]

This channel is like a loophole for math problems.

[@tobybartels8426]

1:37 : Not only is the absolute value not required (when integrating a reciprocal) in the complex numbers, it is *forbidden.*

[@ShapelessMonstrosity]

One great thing about this approach is that you can get a complex definition of arctan and arccot (like how the complex definitions of sin and cos are well-known).

[@AdoNir]

Really cool! That green marker part in the end really made me say “wow” out loud

[@norn-sama3407]

A few months ago I asked a few teachers at my school and no one was able to give me a proper answer if this is correct or not. I thank you so much for this video qwq

[@anshumanagrawal346]

It's not correct, strictly speaking. There's a lot of theory surrounding this, and you can't just do what he did in the video without knowing all that, it's good for having fun, but to make it rigorous you need to know rules about how complex numbers and Calculus works. If you don't believe me, you can notice a few things right off the bat, first the logarithm function is multi valued, second the argument of a+bi is not always equal to tan^-1(b/a), also the identity tan^-1(1/x) = cot^-1(x) only holds for certain values of x, i.e., not always. For example you can see that the original function is perfectly well defined at x=0, but in this way not because we have a 1/x

[@midas-holysmoke7642]

It's correct a think, because the function 1+x^2 is finite and bounded everywhere.

[@kepler4192]

@@anshumanagrawal346 also complex functions works extremely different from normal functions, since they require 4D to work

[@motazfawzi2504]

@@anshumanagrawal346 The arctan(1/0) is undefined yes but with limits it's arctan(infinity) which is π/2 + πn (n is an integer)

[@Xnibblet]

Excellent comprehensive problem that involves many aspects of Calculus. Thank you for showing and explaining this problem!

[@josueramirez7247]

My last college math class was an intro to complex analysis class. I really struggled in that class, but this demo really inspires me.

[@skycraft5447]

J'ai découvert cette chaîne il y a peu de temps et je la trouve vraiment incroyable

[@Jack_Callcott_AU]

Great job BPRP!. It's nice to know partial fraction decomposition works in the complex world.

[@Benthehuman]

What's funny is that this was the first thing I did when I learned how to do partial fractions, but instead of trying to prove that they are similar using polar coordinates (which I had not thought about at all,) I decided to use the taylor series for sin(x), cos(x), and e^x in order to try to set arctan(x) equal to that ln function. I started on Friday and got a decent chunk of the way there but then school ended and I haven't finished. This sort of spoiled it but at least I know it's possible with my other approach now lol (or at least it should be)

[@ragnarokvii]

i always wondered if you could do this! this is amazing!

[@jeremybarrett4535]

Really great explanation. Thank you.

[@jacobperreault6844]

It’s funny, next semester I’m going into calc 1 but I already know how to find the derivative of certain functions and know some basics in integration, love this channel tbh

[@Infinium]

Such a fun video, thanks again! ❤️

[@Gabs2345]

Fascinating stuff! Thank you!

[@procerpat9223]

Beautifully done ! this is pure perfection 👏🏻

[@GaneshGunaji]

Your videos are such a gem! Excellent teacher!

[@debabratakalita9947]

It's a very nice videos. It proves the beauty of complex numbers to it's ultimate level...... Thanks.

[@kiaruna]

Always happy with your great content ! 1M reached soon :)

[@blackpenredpen]

Thank you so much 😀

[@kiaruna]

@@blackpenredpen Omg you replied !!! Hello !!!

[@ahola9338]

Great video !! I was trying to figure out how to go from the imaginary world to the real one and this answered the exact question I had!! Thank you very much for making it ! It’s beautiful.

[@Yuno08888]

This was very amusing and sublime! Thank you so much❤

[@MichaelJamesActually]

You just blew my mind

[@wiwaxiasilver827]

It does reveal a fascinating relationship between ln and arctan.

[@steppindown6874]

Complex analysis is so fun

[@josephtraverso2700]

Absolutely amazing as usual

[@rumapaul275]

well the method was new to me and I really liked it. but one thing in last step where you put tan^(-1)(1/x) = cot^(-1)(x), you should have preferably made two cases x>0 and x0 tan^(-1)(1/x) = cot^(-1)(x) and for x

[@Dreamprism]

7:00 The Green Marker!! :O :O :O

[@conrad5342]

Nice, I remember stumbling across that same question some years ago, but I could not continue from the ln .

[@MohitKumar-eu4pz]

Today after a long period of time ur video made me really happy from inside from these type of question

[@tmogoreanu]

Technically, the transition to exponential form is not correct for cases where x < 0. The angle should be derived differently for (x, i), x < 0 and (x, -i), x < 0. Argument is not just simply arctan(1/x) and arctan(-1/x) correspondingly. I believe there should be an everywhere continuous piecewise function defined first to get a correct primitive and then a proof provided that it's off by a constant with arctan(x). Conclusions in the video are only valid for positive values of x. Please correct me if I'm wrong.

[@JohnnyLaw3134]

Thank you Euler.

[@user-fb2qr4ru6i]

Eu gosto disso! Boa explicação detalhada!

[@rhlogic]

Interesting how the imaginary world intersects the real world in unexpected ways, but still following rules.

[@dalisabe62]

The imaginary world is a bad name for the complex plane. There is nothing imaginary about it. Only because we no better definition for the square root of -1, that we are forced to call it imaginary. We defined the square root of a number in such away that it is the quantity that if multiplied by itself, it would give the number itself. There is no such scenario for negative numbers, and it was a good thing we had none because that prompted the introduction of the complex plane.

[@Batshitcrazy6942]

Haha, I literally did this once out of blu and put limits 0 to pi/4 to find out what e raised to itheta was equal to, I was so happy.

[@ryanpunamiya]

A question similar to this that i did recently was the indefinite integral of (ln x)/(1+x^2). You should definitely try it!

[@hmm.2013]

Can't be done by parts... Very interesting

[@nasirsiddiqui7573]

omfg, this was amazing. in my graduate electromagnetism class we were applying conformal mapping theory to solve electrostatic boundary value problems. there was one particular transformation that involved the logarithms with complex numbers in their arguments and i had no fucking idea how the final answer reduced to some tan^-1. watching you actually go through the algebra has cleared so much up for me. keep at it my guy, college sophomores aren't the only beneficiaries here!

[@diegoc.8518]

u usually know when it's about arctan or arccos or arccsin because depending on your math class u know these are their derivatives. d(arrcos)/dx= -1/sqrt(1-x^2) , d(arcsin)/dx=1/sqrt(1-x^2) and d(arctan)/dx= 1/1+x^2

[@rafaelstv]

The end is brilliant. Without knowing that tan^(-1)(x) + tan^(-1)(1/x) = pi/2, I would have thought that I made something wrong.

[@13579YOOTUBE]

Nice approach sir

[@KJ-on6eg]

Dude I took AP calc ab and this all went over my head but I loved every second of it even though I did not get any of this apart from the first integration

[@Spongebob-lf5dn]

Maybe because they don't teach this in calculus. As per the video title.

[@boodlenorjek]

Lets generalise Productlog(k,(a+bi)e^(c+di))

[@odailgouttai3347]

arctan(1/x)+arctan(x) is equal to pi/2 if x>0 and equal to -pi/2 if x

[@geneyoung11111]

We simply learnt substitute x to tan θ in \int \frac{1}{x^2+1} dx before learning euler formula.

[@KuroiXHF]

I'm a stats guy and this all went right over my head. I still had fun watching, though.

[@felixl3257]

My major is history, i dont know anything he say and this is the most beautiful gibberish for me I have ever heard...

[@wulfrix5967]

when blackpenredpen introduces bluepengreenpen… it’s gonna be awesome

[@General12th]

So good!

[@cherryisripe3165]

Mathematics are indeed exact sciences !!!! Brilliant!

[@dushyanthabandarapalipana5492]

Thanks!

[@franksaved3893]

Can we use this trick for all rational functions? Ex. for 1/(x^4-1) we use the 4-roots of 1.

[@user-sk5zz5cq9y]

yes it works fine for that function too

[@user-sk5zz5cq9y]

but I wouldn't really recommend it because there is better options

[@skylardeslypere9909]

It would generally, but for this example of 1/(x⁴-1) it's particularly easy to see, since the 4th roots of 1 are just 1,-1,i,-i. I.e., you would just decompose it as 1/(x+1), 1/(x-1), 1/(x+i) and 1/(x-i) We already saw that the last two yield the same result as 1/(x²+1) when integrated.

[@ExplosiveBrohoof]

1/(x^4-1) can work exactly the same way, because the fourth roots of 1 are just ±1 and ±i. So you get the same things under integration as you do in this video, plus some extra ln terms. I'd be wary of generalizations though, because you need to be diligent with your choice of constant term and choice of branch sometimes before your solution resolves to a real valued function.

[@Turalcar]

The way he switches markers is kinda mesmerizing

[@repbacchista]

that's a excelent exercise

[@omartorres7249]

I loved the video, hope it will help me in my exam tomorrow.

[@jan-willemreens9010]

...Good day Teacher Steve, How come I can follow your clear presentation very well, but that I would never come up with this in my entire life... Mission impossible for me... Thank you for again another educative presentation, Jan-W

[@brillantstar9308]

Thank you

[@zhangruoran]

You ignored the fact that both ln and arctan are multi-valued functions in complex analysis.

[@kianushmaleki]

It was beautiful

[@chrisrybak4961]

Delightful!

[@jonah1077]

Hey bprp, I'm a high school freshman trying to skip Precalc and Calc AB - everything's in place for me to skip Precalc, but AB is a bit more challenging. Thankfully, your videos make solving these problems so much easier and incredibly intuitive! Thank you so much for your videos, they help me so much. Keep it up!

[@Reallycoolguy1369]

The Khan Academy lessons on Calculus are free and excellent. They are also have questions for mastery. Combine with BPRP and a couple other math for fun channels and you will be set!

[@mr.twicks3009]

All of ab is taught in bc anyways so just skip to bc

[@gabinohernandezgodinez6632]

This is super fun

[@namantenguriya]

Stunned 🥶🤩

[@debabratakalita9947]

Who suggested you the name black pen red pen. How did it came to your mind. Is their any story behind this name. Please make a video on it.....

[@lazaremoanang3116]

Even tough i is not a real numbers so when calculating that area, there's no place for i, when going from 0 to +oo, we'll find π/2 because after integrating we have arctanx and when having (1/2i)ln|(x-i)/(x+i)| we'll have π/2 too. Ok let's watch the video.

[@gasun1274]

do you need an ambulance

[@lazaremoanang3116]

No.

[@Anonim29122]

You could also aply that cosx=(e^ix+e^-ix)/2 and sinx=(e^ix-e^-ix)/2i

[@saitama1830]

isn't that formulat of coshx and sinhx? .. i mean hyperbolic functions

[@affapple3214]

@@saitama1830 yes but you have to just remove all the i's, example: sinh(x) = (e^x-e^-x)/2 That's how I remember sinh formulas, I derive the complex form of sin X and cos X and remove the i's

[@user-vj6ty5lb1l]

@@affapple3214 Cool!! We can also view the connections between the hyperbolic trigs funcs and the complex trigs funcs: e^(ix) = cos(x) + isin(x) (1) e^(-ix) = cos(x) - isin(x) (2) (1)-(2) /2i ⇒ sin(x) = [e^(ix) - e(-ix)]/2i (3) (1)+(2) /2 ⇒ cos(x) = [e^(ix) + e(-ix)]/2 (4) (3) ⇒ sin(ix) = [e^(-x) - e(x)]/2i = i[e^(x) - e(-x)]/2 = isinh(x) (4) ⇒ cos(ix) = [e^(-x)+ e(x)]/2 = [e^(x) + e(-x)]/2 = cosh(x) → sin(ix) = isinh(x) → cos(ix) = cosh(x) In order to remember that we can think sin and cos as odd an even functions. sin(-x) = -sin(x) cos(-x) = cos(x) This isn't just a coincidence. Think about the taylor expantion series.

[@saitama1830]

@@affapple3214 ohh okay mate thanks for the reply. I learnt how to remember that formula in a easy way today ...

[@saitama1830]

@@affapple3214 yeah i was meaning that only.. isn't that a formula of sinhx without i's

[@robsenponte3308]

Very nice man

[@funtimenetwork]

Can you do another video where you solve this using residues? Complex integration

[@VinhNguyen-de7wh]

Arctanx+C

[@anuragyadu2271]

Untying a fine thread demonstrated mathematically.

[@SidneiMV]

AMAZING!

[@teifaf6036]

Wow nice! That is taking the scenic route but it works ha ha

[@poodlebear99]

Awesome!

[@straysoul1301]

Brilliant!!

[@vanshikaramwani9631]

loved it

[@nirajandata]

waiting for your video related to IMO problem

[@hachemimokrane2810]

Vecteur nice method thanks sir

[@rjdnefj6733]

Amazing 👏

[@BridgeBum]

While this interesting it seems like a significantly harder approach than expanding 1/1+x^2 into the convergence of the infinite geometric series sum where t0 is 1 and r is -x^2. That integral and resulting sum is where the (ridiculously slow converging) formula of (1-1/3+1/5-1/7+1/9-1/11...)=pi/4

[@TakeTheRedPill_Now]

imho: The partial fraction substitution step (2nd line) is incorrect as it makes the numerator equal zero. Interestingly the final result is correct, but i suspect when done correctly using integration by parts those 1/x terms in the correct numerator cancel.

[@user-wu8yq1rb9t]

Nice Thank you teacher

[@blackpenredpen]

You are welcome

[@tateconfidential333]

Hey bprp I love ur vids im 16 and I really enjoy calculus :) I have a question for u: could u do integral from -♾ to ♾ of sin(x^2)? Maybe with some Feynman’s I don’t know… Good luck for ur channel!

[@eliw772]

sine is an even function, so that integral is equal to 2 times the integral of sin(x^2) from 0 to inf. To evaluate an improper integral, we have to evaluate the limit of it as it approaches infinity. Sine diverges at infinity, so this integral also diverges.

[@alpcanakaydn6986]

If we write the lower bound of this integral as 0 and the upper bound as infinity we get the equation e^(i×pi)+1=0

[@mikiteti1681]

Awesome video! I wonder if taking the derivative of tan-1(x) is correct as well… It’s easier for me at least

[@srijanabajracharya1086]

Love it

[@hassanniaz7583]

Beautiful

[@maximilianarold]

Everyone gangsta 'till the green pen get's used

[@GSHAPIROY]

I used this to prove Euler's formula without power series.

[@conanedojawa4538]

very good video ☺👍❤

[@danielfetz589]

Very interesting

[@AbouTaim-Lille]

We learned it by heart that it equals: arctg x +C