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

[@blackpenredpen]

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

[@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

[@sambhavkapoor26]

Should've added you know you know too!

[@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.

[@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!

[@RealCrate]

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

[@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.

[@shreyjain3197]

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

[@givrally]

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|

[@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

[@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

[@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

[@tortillajoe]

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.

[@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).

[@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.

[@AdoNir]

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

[@tobybartels8426]

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

[@Sg190th]

I actually forgot about tangent being an odd function. This also helps with re^itheta practice. It all comes back into full circle.

[@GaneshGunaji]

Quite literally full circle. You can also use the unit circle and geometric interpretation of tangent and cotangent to see why the sum of arctan and arccotan is pi/2 as well. It all goes back to the unit circle! :)

[@JayOnDaCob]

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

[@Xnibblet]

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

[@skycraft5447]

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

[@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)

[@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)

[@nayeem7359]

This is such a cool way to do this integral. Really helps you understand complex numbers and their applications

[@militantpacifist4087]

This channel is like a loophole for math problems.

[@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

[@Jack_Callcott_AU]

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

[@steppindown6874]

Complex analysis is so fun

[@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.

[@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.

[@geneyoungdho]

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

[@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.

[@wiwaxiasilver827]

It does reveal a fascinating relationship between ln and arctan.

[@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

[@felixl3257]

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

[@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.

[@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

[@debabratakalita9947]

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

[@MichaelJamesActually]

You just blew my mind

[@jeremybarrett4535]

Really great explanation. Thank you.

[@odailgouttai3347]

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

[@KuroiXHF]

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

[@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

[@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

[@κπυα]

@@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

[@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 .

[@cherryisripe3165]

Mathematics are indeed exact sciences !!!! Brilliant!

[@kiaruna]

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

[@blackpenredpen]

Thank you so much 😀

[@kiaruna]

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

[@MohitKumar-eu4pz]

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

[@Dvir226]

עקרון ההוכחה שarctanx+arccotx=pi/2 הוא הזהות cot(pi\2-x)=tan(x) (אותה ניתן להוכיח מהתבוננות על משולש ישר זווית, זהויות טריגונומטריות ועוד...). עקרון ההוכחה שarccot(1/x)=arctan(x) הוא שtanx=1/cotx.

[@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.

[@GaneshGunaji]

Your videos are such a gem! Excellent teacher!

[@boodlenorjek]

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

[@Turalcar]

The way he switches markers is kinda mesmerizing

[@procerpat9223]

Beautifully done ! this is pure perfection 👏🏻

[@yanntal954]

but arctan(x) and -arctan(1/x) are not a constant apart... only when x>0 and then another constant apart when x

[@JohnnyLaw3134]

Thank you Euler.

[@JSSTyger]

This is my favorite integral of all time. It amazes me that the scary inverse tangent function has such a simplistic looking derivative. I used to prove it the hard way by finding the derivative of the inverse tangent. Tan(y) = [e^(2iy)-1]/[i(e^(2iy)+1)]. Let x = tan(y). x[i(e^(2iy)+1)]=[e^(2iy)-1]. e^(2iy)[ix-1]= -1-ix. e^(2iy) = [-1-ix]/[ix-1]. y = 1/(2i)ln[[-1-ix]/[ix-1]] = tan-1(x). You can then use the quotient rule to calculate the derivative and you get 1/(1+x²)

[@diegoc.8518]

i calculate derivative by (f^-1)'(x)=1/f'(f^-1(x)) in this case (f^-1)'(x)= 1/tan'(arctan(x)) = 1/1+tan(x)^2(arctan(x) = 1/1+x^2 problem solved easily

[@JSSTyger]

@@diegoc.8518 yes ive seen simpler ways to do it. They way i did was just the way i first discovered. And the derivative of that crazy equation being so nice and compact amazes me.

[@diegoc.8518]

@@JSSTyger oh ok nice

[@zhangruoran]

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

[@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.

[@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

[@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.

[@luinerion]

@@dalisabe62 If you think of it, even the negative and irrational numbers have got bad names. We're just so used to it we never notice.

[@omartorres7249]

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

[@wulfrix5967]

when blackpenredpen introduces bluepengreenpen… it’s gonna be awesome

[@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.

[@VinhNguyen-de7wh]

Arctanx+C

[@Infinium]

Such a fun video, thanks again! ❤️

[@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.....

[@Gabs2345]

Fascinating stuff! Thank you!

[@Yuno08888]

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

[@funtimenetwork]

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

[@alejosanchez3309]

If we dont take the branch of Ln that is not defined in the positive numbers or the one that is not defined in the negative numbers the result function is not even continuos so is not an antiderivative.

[@josephtraverso2700]

Absolutely amazing as usual

[@Hyperion1722]

Did you make a shortcut in getting 1/2i and -1/2i? By law of fractions, this should be 1/2x. Or am I mistaken? it is as simple as 1/A + 1/B = (A+B)/AB.

[@ragnarokvii]

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

[@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.

[@hammadsirhindi1320]

Solve for x and y x^2 +y^2 = a x^3+y^3 = b

[@vincentarnould7009]

Hey ! I challenge you with (the best) derivative ever : Find the derivative, with respect to n, of the function that describes the volume of an n-dimensional ball with a radius R d/dn(((pi^(n/2))/Г(n/2 + 1))*R^n) With Г(z) the gamma fonction and R(z) > 0

[@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

[@QuranReact1]

Blackpenredpen, I have an indefinite integral for you : integral of ln(csc(x)+tan(x)).

[@GSHAPIROY]

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

[@niki2669]

Hey, how about dx/[(x+1+sqrt(2x))(x+1-sqrt(2x)) ?

[@datfry7791]

quick question : how do you integrate ln(lnx)?

[@13579YOOTUBE]

Nice approach sir

[@paull2937]

How are you so good at math

[@teifaf6036]

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

[@muktikpatel733]

7:27 that's only when x is positive. The second formula that is.

[@nirajandata]

waiting for your video related to IMO problem

[@kianushmaleki]

It was beautiful

[@karimsaidi8123]

thumbs up if you think we need an ultimate definite integral starter video

[@AbouTaim-Lille]

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

[@SyberMath]

Beautiful solution and nice cover! 😁 I like the following manipulation (Manipulate expressions, not the people!) Note that (x+i)-(x-i)=2i 1/[(x-i)(x+i)]=[(x+i)-(x-i)]/[(x+i)(x-i)]*(1/2i) =(1/2i){[(x+i)/(x+i)(x-i)]-[(x-i)/(x+i)(x-i)]} =(1/2i){[1/(x-i)]-[1/(x+i)]}...

[@cmilkau]

Hmm, this doesn't always work, though, does it? Only analytical functions should have a complex integral, but all piecewise continuous real functions have a real integral.

[@Mernom]

As you saw, in the end all of the complex terms canceled out.

[@anuragyadu2271]

Untying a fine thread demonstrated mathematically.

[@reubenmanzo2054]

I've got a problem for you if you want to try it out: integral: (x+1)/(x^4+4) dx

[@ЯвсеРЕШУ]

Eu gosto disso! Boa explicação detalhada!

[@gergelymorvai979]

I'm not sure if it's really your type of content but you could do ln(x)^log(10,x) = log(10,x)^ln(x) and then all of this equals ln(log(10,x)) = log(10,ln(x)) I did it as a fourteen yo so it probably won't take you long ;-)

[@leolesnjakovic8725]

VERY INFORMATION VIDEO!!! TANK YoU!! (soryr for bad egluish)

[@manasanmol6604]

Well we also need to consider the domain the x must not be equal to 0

[@fept4043]

Dear Steve, I have a further maths exam today, please wish me the best of luck with my chungus integrals and funny matrices. Amen

[@blackpenredpen]

Best of luck!

[@abdullahomran699]

Amen

[@repbacchista]

that's a excelent exercise