you know when someone mastered math when he's watching your video and all he does is the skip five second button the likes and clicks off. i was impressed when I saw him in the library today
Finally, an interal worthy of flexing over my new math friend. imma tell em to sub obviously. I'm so going to use you for next sem's complex analysis to also rizz on my professor, THANKS MATHS 505 KAMAL!
The new audio track feature translating your remarks into other languages is neat. I haven't seen that before. I checked out Spanish and Japanese. The voices sound natural rather than artificially generated. Cool. Expand your audience. Now back to the original from the start.
hi, nice video, i have a few questions where did you study advanced integrals, especially involving the gamma and beta functions? and, where did you do you high school schooling from?
@@rishabhshah8754 I learned alot of math simply in the process of solving integrals 😂 just look for integrals to solve and you should be covered. As far as highschool is concerned, I'm Pakistani and I just did the local high school thing we have here.
How did no one in the comments realize you missed a factor of x squared in the numerator? Pleeeeease it pains me to see a solution development for an integral that is not the one at the start of the video 😭😭😭
The result is correct, but the method of solution is wrong!! You have to avoid the non convergent integral \int_0^\infty 1/(1+x). In order to do that, split the integral \int_0^\infty (x dx)/(1+x^3) as the sum of the integral between 0 and 1 plus the integral between 1 and \infty. Perform the substitution t=1/x in the second integral to show it is equal to \int_0^1 dx/(1+x^3). Thus our integral equals to the arctangent integral (2/3)\int_0^1 ((1+x)dx)/(1+x^3)= (2/3)| \int_0^1dx/(x^2-x+1) which can be easily computed.
Man coughs and is still terribly sorry about that, it's ok man!
Integral :💀
Solution :🥶( chill guy)
guy: terribly sorry about that
Great solution. I’m used to seeing a square root of 2 in an answer; this time it was cool to see a square root of 3 appear.
"How complicated do you want your natural logarithms?"
Maths 505: Yes
Cool math snack🥯
I have a physics exam today, its 2 AM why am I watching this...
Thank you for your innovative ideas.
Hi,
"ok, cool" : 0:40 , 3:33 ,
"terribly sorry about that" : 3:16 , 4:11 , 4:21 , 6:10 , 7:33 , 8:15 .
Favorite parts
The Result is Cool as always
you know when someone mastered math when he's watching your video and all he does is the skip five second button the likes and clicks off. i was impressed when I saw him in the library today
Finally, an interal worthy of flexing over my new math friend. imma tell em to sub obviously. I'm so going to use you for next sem's complex analysis to also rizz on my professor, THANKS MATHS 505 KAMAL!
i really enjoy these vids when i get them recommended lol
The new audio track feature translating your remarks into other languages is neat. I haven't seen that before. I checked out Spanish and Japanese. The voices sound natural rather than artificially generated. Cool. Expand your audience. Now back to the original from the start.
Excellent
Congrats for the job champs 🎉🎉
hi, nice video, i have a few questions
where did you study advanced integrals, especially involving the gamma and beta functions? and,
where did you do you high school schooling from?
@@rishabhshah8754 I learned alot of math simply in the process of solving integrals 😂 just look for integrals to solve and you should be covered.
As far as highschool is concerned, I'm Pakistani and I just did the local high school thing we have here.
Hey broo fan from Sri Lanka ❤🎉
@@sundaresanabishek5127 hey broo friend from Pakistan ❤️🎉
Выражение под логарифмом в числителе представляет собой сумму геометрической прогрессии.
If the alternating sum is replaced by (1+x+x^2...x^2n), wolframalpha says it is not convergent. Is it right?
split interval to (0,1) and (1,infinity), we have same result still
How did no one in the comments realize you missed a factor of x squared in the numerator? Pleeeeease it pains me to see a solution development for an integral that is not the one at the start of the video 😭😭😭
The result is correct, but the method of
solution is wrong!!
You have to avoid the non convergent
integral
\int_0^\infty 1/(1+x).
In order to do that, split the integral \int_0^\infty
(x dx)/(1+x^3) as the sum of the integral between 0 and 1 plus the integral between 1 and \infty. Perform the substitution t=1/x in the second integral to show it is equal to
\int_0^1 dx/(1+x^3). Thus our integral equals to the arctangent integral
(2/3)\int_0^1
((1+x)dx)/(1+x^3)=
(2/3)|
\int_0^1dx/(x^2-x+1)
which can be easily computed.
Where do you see the integral of 1/(1+x) in the video?
Can you be my pookie?
Sire what
Just wanna say hi ;p