fast way to do partial fraction for integrals, calculus 2 tutorial
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- Опубликовано: 2 окт 2024
- why cover-up works: • the cover-up method & ...
integration with partial fraction, calculus 2 tutorial
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You may not read this but i passed my Cal 2 and DE because of you
I read as many comments as possible. Thank you for sharing this great news with me! Good job on your own too!!
I know it's not exact but....
2:57: QUICK MAFS
Angiras Darbha
I don't get this....
It's a lyric from this song by this comedian dude named Big Shaq, Called "Man's not hot" It's filled with absolute nonsense, and it's just hilarious to listen to.
(I feel I got massively trolled)
saw that coming lol
MAN'S NOT HOT 🔥🔥🔥
The ting goes- I mean, the lyrics go :
"Two plus two is four, minus one that's three, quick maths"
It's about what you did in the video at 2:57
Bro u made this a million times easier thank you
My pleasure to help.
first time i've ever seen this method, thanks a lot! even if i dont pass my exams, your channel is excellent and I hope you dont stop
Just learned more than I've done this past semester in calculus. Thanks dude, this is awesome
Great video
Thanks alot...
What if the factors are quadractics or repeated factors?
Have to use equating of coefficients.
2+2=4-1=3
QUICK MAFS...
Andrea Proone
Lol what a nerd..
You don't get it ?
Andrea Proone
-__-
WHAT THE HECK IS THIS MAGIC
You are a saint!
daddy i love you u saved my life
Super helpful!
2:54 Two times two is four, minus one that's three. QUICK MATHS.
OMG.
That was really helpful!
Thanks!
thnx a lot
Wuuhh great sirrrr vry nycc solution I like it....👍👍👍
That's good!
THANKYOU!!!
wonderful video, this is gonna save me on precal test tomorrow
Awesome that was helpful
Nice, liked it! :D
1:28 why cover it up when solving x-1=0?
Ty
I never seen this technique in our school
Good teacher. I'm ateacher of math all so.
Excellent job on your explanation as to how the A, B, and C are derived!!!
It helps a lot .👍👍👍 Thank you sir
Would you use partial factor also in case of differentiating the same equation?
How to solve the following definite integral:
1/((1+x^2015)(1+x^2)) from zero to infinity, plz help me on that...
Wouldn't you just expand out the bottom, then you just have 1/some polynomial, then you differentiate that polynomial and times it by the natural log of the polynomisl
yeah youre right, i guess you would have to split up the fraction and then integrate normally?
im sure its way above my head so dont bother ahah. I would however want to know why you cant split the fraction and integrate the answer, if the explanation isnt too hard :)
ThePolm3 thanks for the info, yes i am interested in knowing how you derive that answer. Pi/4
ThePolm3 , Thank so much Sir for the neatness and clear step by step solution , Most appreciated 🙏🏻👍🏻
Kindly
Solve by partial fraction
(4x^4+6x-7)÷(x^3+x-1)
❤️
Finally a useful life hack video on RUclips!
What if there is A + Bx +C in the numerator????
How about that x??
anyone here that could verify that this is a correct method like has anyone used this and it got marked correct?
So what’s the answer written as one ln, since you can simplify it using the rules of logarithms.
Help me! @blackpenredpen solve this integral (5x^2+3x-1)/((x-2)(x^2+1)), How this problem is solved by your method?, What to do the quadratic term?, Can i use complex numbers?
Omg you fucking saved my exam xD
You are just BRILLIANT - THANK YOU
Why wasn't I taught that in college?
Integrate (x³/(x-1)(x-2)(x-3))dx. According to this method, the next step is 1/2(x-1) - 8/(x-2) + 27/2(x-3). But the actual partial fraction will have a + 1 added to it. Why is that so? Where am I going wrong?
Thank you so much! 😘 You're the best teacher 😍👏👏
So what are it's limitation?
Can we prove it that, why happens so??? ...
It was good and I did understood
You are the best
This is quick math
Dude this guy is legendary
❤
OMG
0 u sub :)
Man you're amazing finally I understood PFD
But you can do loads of work to simplify the answer
Hey, I'm sorry, I've been through multivariable and for some reason I seem to have forgotten a few of the basics. Why is it that we have the absolute value inside of the logarithms that result from integrated inverses?
Because the domain of log functions is only the positive numbers. Consider y = ln(x). That is the same as e^y = e^ln(x) and then e^y = x. For any real y, x can not be negative. So to prevent an undefined value sneaking in we apply absolute value to the input of the log.
Ah, I see. Thank you!
it was good but
thu thu kar di sari jagah
You are such a genius dude not even gonna lie
what if the denominators has duplicate (repeated ) linear factors? That technique is applicable also?
Tbh I think I would just be quicker just doing it the normal way. Not quickly than you obviously lol just me personally
Hey, well I would ask that how could you change those different white board markers so quick, can you plz tell me , it will be helpful in my exmas where blue black creates a lot of mistakes
it just shocked me to see that partial fraction decomposition is actually the same as using cauchy's integral theorem oO
How is that?
From what I'm aware of, the theorem simply states that if you have a function f(z) which is holomorphic between two paths that it is therefore equivalent to take the contour integral along any path on or within the region.
Not entirely sure where it connects to partial fractions.
ok, it would more precisely be called cauchy's integral formula. in a sense, like the residue theorem.
what if there are (x-a) and (x-a)^2?
use long division and solve from there. Your remainder becomes the new numerator of the original function and the quotient becomes its separate integral. Use u-sub if necessary
Can u do this on all partial fraction problems
What if in the denominator there's for example (x-1)^2? Does the cube change the way the solution is done ?
to short video !!!
boooooooooooooooo!!!!!!
; D
now this is calculus :(
Oh wow this helped tremendously
It's amazing and easier.Thank you.
does it work for positive numbers
Man you the best! You deserve all the love in the world.
This video needs more views. Really helpful.
thank you sir very very much.
Very helpful cool to see it done this way thanks!
+ C (not to confuse to earlier C=5/2)
Thank you so much.You are really a good math teacher.😊
Can we do the same for non linear or other order fractions
ur the best man! Thank you :)
Thank you! This was so helpful!!
Best channel on RUclips 🖤❤
Just loved it. ❤
Thanks
What is going on!!???
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I love you
Does that only work for the linear numerator? As in, suppose my numerator is a quadratic and my denominator is a linear and a difference of 2 squares, would this method work?
yes it would. As long as a standard procedure can be used to separate the compound fraction into integrable partial fractions
Dude, i love you
life hacks
For multiple roots we can use this approach
Int(P(x)/Q(x)dx)=P_{1}(x)/Q_{1}(x)+Int(P_{2}(x)/Q_{2}(x)dx)
Q_{1}(x)=GCD(Q(x),Q'(x))
Q(x)=Q_{1}(x)Q_{2}(x)
deg P_{1}(x)< deg Q_{1}(x)
deg P_{2}(x)< deg Q_{2}(x)
Polynomials P_{1}(x) and P_{2}(x) we can get from undetermined coefficients
fantastic
Does the answer to the integral of (1/2)/(x-1) also equal (ln(2x-2))/2? I worked it out by multiplying the fraction instead of removing the coefficient from the integration. I think it gives me the same answer (but with a different value for +C) but I'm not sure... If I expand it, I get (ln(x-1))/2 + (ln(2))/2 +C. Please could you tell me if this is the same?
Yes it's the same, ln(2)/2 is just a constant. So you can call ln(2)/2 +C=C' and it's the same.
Legend :)
Hi what if we he two coefficients in the numerator like
A/(S+3) + (BS+C)/(S^2 + 4) = (6S^2 + 50) / ((S^2 + 4) (S + 3))
www.snapxam.com/problems/79162013/integral-of-2x-1-0-x-1-0-x-2-0-x-3-0-dx
#blackpenredpen
can you please help me in solving integral of:.
(x^5-x^4+4x^3-4x^2+8x-4)/(x^2+2)^3
integral(x^5 - x^4 + 4 x^3 - 4 x^2 + 8 x - 4)/(x^2 + 2)^3 dx = 1/2 (-2/(x^2 + 2)^2 + log(x^2 + 2) - sqrt(2) tan^(-1)(x/sqrt(2))) + constant
Thank you 💓💗💖💔💖💔💞💓💕💘💘💘💘💘💘
cykaty tiukplos wow
How could be done Asti su of function of the kind y=f(x)-g(x)
Y= ln|x^2-5x+6| - (x^3+2x^2-X-1) \(x^2-4) ?
Yeah but why does this work
www.snapxam.com/problems/79162013/integral-of-2x-1-0-x-1-0-x-2-0-x-3-0-dx
Why does that work?
If you simplify it back into one single fraction, you get the original
Easy!!! You made Sal Khan sucks!!!
It is easier to do equating coefficients
Does this work for same factors? Like for example, if I have A/x-1 + B/x-1
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see, i can read your mind!
omg, you are god! this is the proof that god is a mathematican! thank you :D
@@blackpenredpen You could've just said no it doesn't work lol
This is gonna save me like 20 seconds per question and it might not look like it but that's a lot in competitive exams!!! You made my day
I read it you made me cry
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