I am very glad to hear about this! And I plan to make more of this kind of integral videos. I think it's a really good way to really learn the integration techniques and more fun too!
This is maybe my favorite youtube channel. I love the way you approach the different problems. It teaches me not only how to solve the problem, but also how to approach other problems. Thank you! 谢谢！
This sort of thing reminds me of some other methods that often tutors seem to not know when approaching what can be very easy integrals like integral xsqrt(1+x) dx doesn't have to be by parts instead write x=x+1-1
Wow!Great videos,great explaining and great examples!Thank you so much! Also I have some integrals examples which I am dealing with lately and I am having some diffuculties to deal with!Would u be able to have a look to them and help me with? THANK YOU SO MUCH!
Well i came to the same answer but i used the method of partial fraction decomposition and did the integrals subsequently. This method needed a lot more work than yours.
I did the first one by adding and subtracting x^3 in the numerator 1dx / x(x^3+1) = (x^3+1-x^3)dx / x(x^3+1) So I got a sum of two integrals and the answer I got is ln |x| - 1/3 ln | x^3+1 |, and it's equivalent to yours :)
es interesante que la forma en que resuelves la primera te simplifica bastante, yo lo primero que pensé fue por fracciones parciales y si llegue a un resultado equivalente al tuyo, pero por parciales es más largo el procedimiento, en fin excelente vídeo saludos.
wow i solve this using trig sub which took me 1 page.. and this method of yours is extremely useful and easy. have you come up with a name for this method yet?? heheh
but the fourth root of x raised to the fourth is x only if x is greater than or equal to zero, otherwise it would be -x .... I have that doubt, then we would have to make two different integrals, one assuming that x is positive and the other where the x is negative? help me please!!! PD: excuse my english but i speak spanish
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This is just ingenious
.
Asaf Shahar thank you
I watch your videos all the time and learning so much more than what the cal class offers, Great video.
I am very glad to hear about this!
And I plan to make more of this kind of integral videos. I think it's a really good way to really learn the integration techniques and more fun too!
Dude, I stopped watching South Park for fun and started doing integrals. You rock!
I did substitution of tanθ=x^3/2 but your way is much nicer. Great video ;)
*Edit: For the first integral that is*
It would be great if you make a video explaining hyperbolic trigonometric functions.
This is maybe my favorite youtube channel. I love the way you approach the different problems. It teaches me not only how to solve the problem, but also how to approach other problems. Thank you! 谢谢！
Cole Hudson thank you!
Thanks for making all these awesome math videos for us math people around the world! :)
Another way to do the first one: 1/(x^4 + x) = 1/x(x^3 + 1) = 1/x - x^2/(1+x^3) and antiderivative is just ln x - ln(1 + x^3) / 3 + C
Alan Yan that's exactly how I did it as soon as I saw it.
This sort of thing reminds me of some other methods that often tutors seem to not know when approaching what can be very easy integrals like integral xsqrt(1+x) dx doesn't have to be by parts instead write x=x+1-1
This is art
Mind blowing strat! Thank u ❤
Excellents tutorials, I've learned more and more, practice and love for maths, two essential things, thanks for all your wonderful posts.
Wow!Great videos,great explaining and great examples!Thank you so much!
Also I have some integrals examples which I am dealing with lately and I am having some diffuculties to deal with!Would u be able to have a look to them and help me with?
THANK YOU SO MUCH!
wow, i never knew there was such a way to solve an integral, thanks for sharing!
Yea, this only works for special occasions
Visit our channel to find great integrals.
I take my words.You are not only good mind manupulation, you are "PRO" at this🤯🤯🤯😎😎.I felt sorry bcoz I have shortage of words tp explain it.
Eres el puto amo!!
i did the first with the separate fractions method and then using the exact approach showed for the second one.
solve integration (x^2+1)(x^2+2)/(x^2+3)/(x^2+4)
good problem, maybe i will out it out in a vid one day
1/3 (3x - 9 tan^-1 (x/2) + 2 sqr3 tan^-1( x/sqr3)) + C
Visit our channel to find great integrals.
Correct, i had the same answer.
Well i came to the same answer but i used the method of partial fraction decomposition and did the integrals subsequently. This method needed a lot more work than yours.
thank you. this video is good
I did the first one by adding and subtracting x^3 in the numerator
1dx / x(x^3+1) = (x^3+1-x^3)dx / x(x^3+1)
So I got a sum of two integrals and the answer I got is ln |x| - 1/3 ln | x^3+1 |, and it's equivalent to yours :)
Integrating of binomial differentials
In polish and russian schools there is standard substitution for this
es interesante que la forma en que resuelves la primera te simplifica bastante, yo lo primero que pensé fue por fracciones parciales y si llegue a un resultado equivalente al tuyo, pero por parciales es más largo el procedimiento, en fin excelente vídeo saludos.
Great piece of work
in the first one, I factored out x^2 and made two subs. It worked
wow!I got it... I had no idea there was such a way!thank u.
great man...u r the boss
wow i solve this using trig sub which took me 1 page.. and this method of yours is extremely useful and easy. have you come up with a name for this method yet?? heheh
Great !!
that was actually fun! specially because I didnt have to solve them
The video was very helpful.
Thanks
great work i love it
Visit our channel to find great integrals.
what does he mean @1:20?
Thank you.
Very nice
I have made both of them maaan!!!!
Thank you :)
Genius
1/(x^8 + 1) ?????????
Never would i think so.
At 8:33 the divided by 3 has vanished. Great way to do it took me a while doing it the partial fractions way.
Alex Goodwin I see. I will have to edit that out
You forgot to multiply the integral by sign(x).
How can you factor out an x^4 power in the denominator of the second integral if the quantity you are factoring it out of is raised to a power?
excelente, saludos desde Colombia
These type of integrals simply solved by taking x power in common
what if we do tan substitution on No2?
How to do this math
integrate
(1-v/v२+v+1 ) =1/x dv/dx
thanku very good way
Para la primera integral sume un cero: 4x^3-4x^3 y sale directo
Nice teaching :)
Thanks for thes vedio: )
You're welcome. I am glad that you like them!
In india ,we call it 'kutur putur ' method.thanks 😍
but the fourth root of x raised to the fourth is x only if x is greater than or equal to zero, otherwise it would be -x .... I have that doubt, then we would have to make two different integrals, one assuming that x is positive and the other where the x is negative?
help me please!!! PD:
excuse my english but i speak spanish
Te amo :)
Do you always have to make calculus this easy 😂😂😂😂
(1+x^3)
why not split 1st one as 1/x - x^2/(1+x^3) but answer is different from yours.
wadbor wahlang lnx-ln(1+x^3)/3= (ln(x^3)-ln(1+x^3))/3 = -ln((1+x^3)/x^3)/3 = -ln(1+x^-3)/3; as you can see, both answers are the same.
genius
thanks!
dude, your mic is just ridiculous
Sir plese integrate 1/(x^7+ 1)
Well this integral can be done by complex numbers.
please solve that...... dx/x^2(x^2+1)^2
You can solve this integral 2 ways: trig sub and partial fractions, try both, by the 2nd one you need integration by parts.
Easy 👌
1/(x^4+x)=1/x-x^2/(x^3+1) , the rest of steps are clear.....
(x^-3 + 1) ???? negative -3
pinche nonito donaire
?
Thank you.
Thank you.