What Integration Technique Do I Use? Example 1

[@ringo3650]

People ended up here, either proving that their school teachers are being lousy, or they don't pay attention during the lecture. Or perhaps simply that Patrick is cool.

[@dextermorgan9609]

. 念來過反會才頭木 All of the above for me my friend.

[@hurrdurr25]

+. 念來過反會才頭木 more likely the teachers are lousy. people that don't pay attention probably wouldn't care to learn how to do the work and never make it to these videos.

[@thewiseone9798]

+. 念來過反會才頭木 Why do you have Chinese characters in your name?

[@patrickjmt]

+. 念來過反會才頭木 i think the correct answer is: all of the above (especially the last part)

[@ConnorJW96]

+. 念來過反會才頭木 Integration and differentiation are always good things to practice, as they crop up all the time. I find myself coming back to this video (and others by PatricksJMT's) to refresh my memory. Very useful!

[@jackmathieson1903]

In my opinion, knowing which method of integration to use is the hardest part of A Level Maths. I've got my Pure exam tomorrow.

[@mileywaffle]

Thank you SO MUCH. I've been subbed and watching your videos for 2 years now. I really appreciate that you keep it free and accessible.

[@douglasleigh2831]

The

[@Agreedtodisagree]

I can't emphasize this enough, practice practice and practice. If you think you got it, then choose a hard problem and then try it. And kudos to this guy for having such a love for math.

[@ahmadzaka9885]

Not even kidding, this video probably just saved my life, you have no idea how much this helped me.

[@ThePharphis]

After a few years of calc I'm feeling pretty confident about most integrations, but I still think this video is helpful.

[@intelliza7248]

Write the numerator as (1-e^x) +2.e^x and then split it in two parts ,integration of first part is x and to integrate the second part substitute u=(1-e^x)

[@212ntruesdale]

BINGO!!!!!!!

[@CheapPhysics]

Bingoooooo.....

[@johnman5391]

Hey, I'm still learning this stuff can you explain that more simply? Is that method simpler?

[@kelbyreyes3599]

why do i always have a feeling that i'm doing everything wrong :(

[@finback2005]

+Kelbypursuer ' it not just you

[@markoplahuta8284]

because you are lazy and think everything will just magically look solvable xD So you give up as soon as something starts looking more complicated than it was. First thing, you see people like patrick solve what they have either already solved or they have a lot of experience in solving and of course that it is gonna look like it's supposed to be easy. Secondly, most important thing is to know the basics. Yes, you can overcomplicate a problem to an extent at which you should start over or look up the answer, BUT, be sure that you have at least gotten to that point by using correct "mathematics" (aka didn't miss a minus, didn't do something that you can't with square roots etc. Basically don't integrate some function, before you know what that function is, how it behaves, how its plot looks, what can you do with it with some algebra . And do keep going, even when it looks like next step will require too much effort to do. You slowly train that way

[@vanessasarahilizarragaoliv6591]

hey nono, don´t feel like that, you must trust yourself, you´re learning so you are fixing your mistakes, it´s ok while you are understanding what you do

[@DylanM333]

because calculus is a deep ocean of knowledge and you literally build billions of brain pathways to be successful.

[@jacobbchapman5644]

Integrals are hard.

[@hari8568]

simply divide num and denom by e^x then we get e^-x+1 divide by e^-x -1 then add to num +1-1 and split integral we get integral e^-x /e^-x -1 + integral e^x /1-e^x.solving by substituting for both integrals by t=e^-x -1 andt=1-e^x in the end we get -log(e^-x -1)-log(1-e^x)

[@patrickjmt]

yes, i will make one eventually ! it would be a great one to make, i agree

[@ladicius]

as a current student in differential equations, i have one bit of advice for you: try to find the staedtler liquid point 7 pens (that are actually a 0.4mm tip). as a user of the v5's for many semesters, the switch was absolutely fantastic for me. it's worth giving them a test run. thanks for the videos. i like watching them over breakfast some mornings.

[@nickthorpe7567]

I split the numerator up into 1-ex+2ex so that I could split the fraction up into int[ 2ex/(1-ex) + (1-ex)/(1-ex) ]dx Then you get left with Int[2ex/(1-ex) + 1]dx Int[2ex/(1-ex)]dx + x Let u=1-ex Let du=-exdx -Int[2/u]du + x 2ln|u| + x 2ln[1-ex] + x + c Little bit more intuitive to me imo!

[@JirivandenAssem]

Yep for me 2

[@Peter_1986]

Awesome, integration is just what I need to practice for my exam in Linear Algebra & Integral Calculus. Thanks a lot.

[@patrickjmt]

and mainly, this is how i did it :) i do not claim it is the fastest, but i do claim it works.

[@dreia2405]

you made this very easy integral complicated, just add e^x and subtract e^x from de numerator

[@executorarktanis2323]

the first substitution was a eye opener

[@kkamous7278]

yes I feel happy from your reply . I get lots of help from your integration videos. you're so genuine . tomorrow I have presentation in university and I got lots oh help from your video s.

[@BubbleGumCMM]

i wanted a video for this! nice timing. THANKS PATRICK!

[@Monadoabyss]

A quicker way: (1+e^x)/(1-e^x) is just -coth(x/2) which can be easily integrated as it's in the form f'(x)/f(x).

[@flupypersonyui3488]

That's gold, Jerry! Gold!... I mean Patrick.

[@patrickjmt]

it does not have an elementary antiderivative ; you have to use series to do this

[@marcusrayrosales1]

This is way easier than shown here! First, multiply top and bottom by e^{-x/2}, so that: (1+e^x)/(1-e^x) becomes: -(e^{x/2}+e^{-x/2})/(e^{x/2}-e^{-x/2}) ==-cosh (x/2)/sinh (x/2) The integral is easily seen to be: -2ln [sinh (x/2)]■ knowledge of hyperbolic sines and cosines are not necessary here; just do the first step of multiplying by e^{-x/2}, then let u be the denominator. All so much easier than this video shows.

[@jackmcclane6113]

I think you need to know that he’s not solving maths for himself, he is showing it to learners in details no need for a short way!!! When students understood the detailed procedures, then short way is now great.

[@patrickjmt]

the second you can do with a u sub, but you still need to do the same type of substitution i did (i think!) for the first one

[@billwindsor4224]

Great explanations here, Patrick. I purchased the worksheet -- it's excellent (and inexpensive); I encourage you to develop more worksheets, and to give them more visibility in your video posts!

[@mohanarora1]

You are amazing dude . Became a die hard fan of your s after this video

[@ghostkr3676]

Thanks man..... I always wondered how to start..... and you cleared my doubt

[@satrickptar6265]

I'm thanking my advanced algebra teacher in grade 9 ;-; it really helped me in integral calculus in grade 10!

[@baburo101]

Finally, Pat's saving some trees!

[@fergusfs70]

In your final answer you leave ln(e^x) which can just be written as x

[@blackham7]

But it helps us students see what's going on. He might know that but we might miss that.

[@hamid-we7vs]

At 4:40 , could you have used the reverse chain rule after doing the u-substitution ?

[@edgaracosta9976]

I'd just like to comment that there is a way to at least know how to approach one type of integration problem. If you are multiplying or dividing two functions of a different type, the only approach is integration by parts. For example, finding the integral of (x^2 + 2)(sinx) dx. This type of problem would be impossible to do through substitution because in the most general cases, you need the functions to be of the same type, that way you have a chance of figuring out which one is the derivative of which.

[@br5757]

You are a wonderful teacher!

[@Darwinion]

Can that final answer be simplified to: x - 2ln(1-e^x) + c..... as ln (e^x) = x ???

[@jmadluck]

I thought the same but I'm not sure if the absolute value messes with that

[@lich2012]

Hi friend. Good video. Could you explain the method for integrals with several sin and cos founctions where x must be changed for tan(t/2) (or something like that, i dont remember clear right now) please?

[@soupisfornoobs4081]

I believe that's called weierstrass substitution, you can look it up and get some interesting insights

[@JirivandenAssem]

Its the halfangle tangent substitution. Pretty easy, just check out wikipedia

[@amak1131]

This is indeed the hard part of Calc 2. Actual problem aside (my Calc 2 professor gave crazy integrals to do), the hard part is which method to use.

[@yastronaut3]

Hows life now? What are you up to?

[@mayahmed5476]

Your the best. I have a final tomorrow and ur video helped me alot

[@prateekgurjar1651]

you're*

[@yashwanthandaluri8934]

sollu

[@212ntruesdale]

You made a challenge out of a very easy integral if you make one very clever substitution. Instead of 1-e^x, make the numerator 1-e^x+2e^x. Then you have two integrals, intdx + 2int[e^x/1-e^x]. Couldn't be simpler!

[@japori2000]

Yes yes yes. Just the video I needed!!

[@iBoom3RxX]

At 6:34, where did the denominator of -1 come from?

[@LoneWolfKTG]

From the second u-sub that he shows immediately after...

[@iBoom3RxX]

Thanks mate!

[@LoneWolfKTG]

Ginzaki no prob!

[@LoneWolfKTG]

Ginzaki oh and Dark Souls ftw !

[@matblitz4637]

you. can also use substitute x =-t and then do it

[@patrickjmt]

age and allergies and new microphones all do that

[@asiyaishal]

what if we take e^[x/2] out from both denominator & numerator..Then substitute the whole dnmtr term with 'u'....to get a integral [1/u]du form... #justAsking #student

[@and_still.]

I just subscribed you. Thanx from India.

[@shashwatbajpai1350]

You could have multiplied numerator and denominator by e^(-x/2) then you could have directly substituted your denominator as u then your numerator would have been - du and you would have got the answer in 3 steps

[@seroujghazarian6343]

7:47 you can't be serious! You should know that ln(e^x)=x, and e^x is always positive for all real x, so you can't tell me it is in absolute value because abs(e^x)=e^x no matter which value x has.

[@HotPepperLala]

Can you do a video on Euler_substitution? And possibly Elliptic Integrals?

[@crazyphil7782]

The easiest way to do this is using hyperbolic identies. 1 + e^x = e^(x/2) [ e^(x/2) + e^-(x/2) ] 1 - e^x = e^(x/2) [ e^-(x/2) - e^(x/2) ] So it is basically -1/tanh(x), and integrating that is child's play.

[@ComandaKronikk]

dude thats what i thought immediately!

[@ComandaKronikk]

still really helpful video

[@nascar12141]

I wish I had your awesome penmanship

[@kkamous7278]

you are the great man. I love your work. you work for human. I love u so much. alas I meet with you . . I watch your video's. I think you are the best man in world of mathematics. May you live long. my prayer with you.

[@patrickjmt]

+Muhammad Kamran Khan thanks for the kind words, I appreciate it :)

[@akarshansaraf721]

You could have easily done by dividing numerator and denominator by e^x/2 and then substituting e^x/2+e^-x/2=u

[@Chemicalsworld76]

Poti you work very good

[@cypher4528]

Hey man, the question (1+e^x)/(1-e^x) Is it OK if I possibly solved it by splitting it into 1/(1-e^x)+e^x/(1-e^x) Where I multiply (e^-x) to numerator and denominator of 1/(1-e^x) And gave (1-e^x)=t and e^x dx= -dt to e^x/(1-e^x)?

[@damianflett6360]

No. I do not permit it

[@guitarttimman]

Very good.

[@hadireg]

great job! thanks!

[@patrickjmt]

nope, never done anything at all with them.

[@bcarray]

what kind of pen is that? i had a same one and loved it but i can't find one anymore :'(

[@pvic6959]

+bcarray www.amazon.com/s/ref=nb_sb_noss?url=search-alias%3Daps&field-keywords=precise+7+pen that might be of help

[@Andr295]

You rock bro

[@wyboo2019]

im wondering if you could make it a trig integral by letting x = iu, and then e^x = cos(u) + i sin(u), but im not sure too what extent this is mathematically rigorous

[@JirivandenAssem]

Brother always see if u can write the numerator as the denominator. I immedoately see that u can add 2ex to the nominator to get 1-ex+2ex. Split integral and notice one is x and one is ln(1-ex) cos derivative of ex is ex In the end, a usub is not even needed !

[@JC-cu2ym]

@patrickJMT cant we use Quotient rule to solve this example?? Instead of substitution?

[@OUSSA]

There is a much easier way in to write the top as 1-exp(x)+2exp(x) to simplify the integral and get 1+2(exp(x))/(1-exp(x))

[@cypher4528]

I only noticed that you are Left Handed after watching 10 of your videos.

[@patrickjmt]

then what

[@sandyshores8921]

thank you.

[@peregrinefal6607]

Why not split the first problem to: 1/(1-e^x) + e^x/(1-e^x) ?

[@sophiacarryl]

Do you have a video on the second theorem of calculus?

[@EverydayXperiment]

I wish I was this good!

[@gozao100]

i have a question is it normal a function to have more then one integral form?i mean a did this exercise and the result was x-2log(1-e^x)+c and it's the same result when i derive this or am i doing something wrong?

[@dannyawesome63]

I took Calculus at UC Davis back in 2015... and man.. I don’t remember any of this LMAOO

[@DrHansPeterWurst]

that one is really fun...

[@rsmith5645]

How do I integrate 1/(x^1/3+x^1/4)+ln(1+x^1/6)/(x^1/3+x^1/2)dx?

[@damianflett6360]

Do you still need to know

[@JS-ns8dr]

@@damianflett6360 LMAOOO

[@damianflett6360]

@@JS-ns8dr can’t believe they never got back to me.

[@venkateshmangrulkar4776]

In denominator quadratic is compulsory in partial fraction thats why i don't understand why u solve this by partial fraction

[@PranayMundra]

You could add and subtract e^x in the numerator and then take 1-e^x/1-e^x as one integral and the other as 2e^x/1-e^x where the substitution looks more familiar.

[@lezero211]

Hmmm can't seem to understand the part where you used ln to solve for dx

[@kyler2532]

well it is the relationship between exponents and logarithms. When you multiply ln to e^x ( ln(e^x) ) you will get x. Also in order to make the equals sign true he multiplied the other side with ln. His way is much easier than trying to solve the integral when both u and du=e^x

[@zhubarb]

Thank you, this is elementary and useful information.

[@veggiekeller]

For the problem finding the integral of (1 +e^x)/(1-e^x) why couldn't you just break up the fraction at the beginning and integrate 2 fractions (1/1-e^x) + (e^x/1-e^x)

[@HamzaDjeloud]

This is a very easy way Thank you

[@kkamous7278]

I am really feel so happy because you reply me

[@patrickjmt]

ha, glad i could make you happy :)

[@cepson]

It's weird to me that this problem can be solved by letting u = 0, when u is really e to the x and can't be 0. Yet it works.

[@jacobbchapman5644]

What's the difference between an Integral and an improper Integral?

[@OUSSA]

Improper means it has an infinity in one of its boundaries or both

[@chasegregory1328]

For every percent I get over 50% on my exam tomorrow, I'll donate $1.

[@blackham7]

Can I have the money

[@damianflett6360]

How much did you donate

[@algebrayd]

are you the same person who does the cengagebrain solution videos??

[@beta5770]

CHALLENGE ACCEPTED

[@ch0vits]

gonna watch the partial fractions to understand this:)

[@shahriarfardin777]

they are o level math how you can hope to watch this without learning those cuz integration is a level one

[@mengnjoeusebia1335]

Please could you just create a mnemonic for that method

[@mengnjoeusebia1335]

I got it "kuit a trip"

[@TheHateraboi]

can you do a video on e^(pi i) = -1??

[@idiosyncraticidiott5015]

Cuz we all recognize -coth(x/2) when we see it

[@sineadh1994]

where did the absolute value signs come from?? i understood it until there

[@Peter_1986]

The absolute value signs make the natural logarithm more general, so that it is valid for negative value of u as well. If you try comparing ln(u) and ln|u| on a graphing calculator then you can see this more clearly.

[@lucasm4299]

When we have 1/x and take the antiderivative we have to take into account the domain of the antiderivative. Derivative Ln(x) - 1/x The ln(x) domain is x>0 The values of 1/x respect the domain Antiderivative 1/x - Ln(x) The domain of 1/x is all numbers except 0, ln(x) can't take negative numbers so we have to find a way to extend its domain. |x| solves that problem. If you were wondering, d/dx ln|x| = 1/x

[@theBang4thebuck]

I’m in calc 2 and I’m about to throw in the towel

[@kkamous7278]

thanks to my God that you read my comment. l love you sir so much.

[@murugesankgm1297]

you could do it by rationalising it and resolving

[@wolfgangi]

I fucking hate integration by parts especially my memory is wiped out by my crippling alcoholism lol

[@alexgabel4379]

please try to solve ∫e^(-x^2) dx. There's a way to solve it using a 2nd dimension (by squaring the whole thing) and then you get: ∫∫e^(-x^2-y^2) dxdy, but then I get stuck...

[@realitycheck816]

Lol you could have just divided the numerator and denominator by e^x. Then it becomes an easy log integration. 3 lines maximum lmao

[@Grantbraner]

Hi Mr. Woody

[@fastchill]

an other option you should consider is partial integration.

[@sabithasasikumar239]

Nooooooooooooo.....I discovered patrickJMT through google ;)

[@patrickjmt]

ah, in that case, i will just delete all my videos ;)

[@baharosman1416]

Derivatives are more lovely than integrations 😅

[@JS-ns8dr]

hmm yeah true

[@someshrao1087]

sir, please give practice sheet free of cost ..please.

[@ghost880106]

I love the pens that you use. Just saying. Haha