... First of all Newton, I wish you a healthy, peaceful and education rich 2023! That you can make our world again a little smarter in your characteristic Newton way! Regarding this indefinite integral I see myself sitting in the middle of a vast desert (Sahara) without any tools (no textbooks, etc...), with only some basic calculus experience and common sense, and a lot of valuable sand to write on. This integral invites me to solve it with simple means (lol). No abstractions (u-subs etc...), maybe also to take away the general fear (its abstract character) that is going around for maths in general. I'm still enjoying some free time, but wanted to wish you all the best for 2023 from Europe, and hope to have some interesting (math) discussions for the coming future with you. Your presentation was very clear as usual ... Thank you and take care Newton, Jan-W p.s. I would start investigating your integral by differentiating 2^(sqrt(x)), and see what will happen (lol) ...
@@PrimeNewtons Re-express and substitute - 2 ∫ e^(ln2√x) / (2√x) dx u = √x, du = 1 / (2√x) dx 2 ∫ e^(ln2u) du = (2 / ln2) e^(ln2u) + c = (2 / ln2) 2^√x + c You can be the judge if it's easier, I'm biased, I go to the e world whenever I can. Often it's more work but not the kind I mind. 🤷🏻♂️
You are the best math-teacher on youtube! Brilliant language and a very tranparent explanation.
Glad you think so!
... First of all Newton, I wish you a healthy, peaceful and education rich 2023! That you can make our world again a little smarter in your characteristic Newton way! Regarding this indefinite integral I see myself sitting in the middle of a vast desert (Sahara) without any tools (no textbooks, etc...), with only some basic calculus experience and common sense, and a lot of valuable sand to write on. This integral invites me to solve it with simple means (lol). No abstractions (u-subs etc...), maybe also to take away the general fear (its abstract character) that is going around for maths in general. I'm still enjoying some free time, but wanted to wish you all the best for 2023 from Europe, and hope to have some interesting (math) discussions for the coming future with you. Your presentation was very clear as usual ... Thank you and take care Newton, Jan-W p.s. I would start investigating your integral by differentiating 2^(sqrt(x)), and see what will happen (lol) ...
🌞😂👌🎆
Thanks Jan-w. Happy new year to you too. Let's hope for the best. And mor3 learning too 😂
Woow this is so wonderful, where were you when I was doing advanced calculas??
Lol. I was learning 😂
First comment very helpful
Can upload some videos solving differential equations with constant coefficient
I'll think about it
@@PrimeNewtons thank u
alternatively would it not be easier making sqrt x the u?
I think it would take more substitutions. Try it and give me feedback, please.
@@PrimeNewtons Re-express and substitute -
2 ∫ e^(ln2√x) / (2√x) dx
u = √x, du = 1 / (2√x) dx
2 ∫ e^(ln2u) du
= (2 / ln2) e^(ln2u) + c
= (2 / ln2) 2^√x + c
You can be the judge if it's easier, I'm biased, I go to the e world whenever I can. Often it's more work but not the kind I mind. 🤷🏻♂️
Happy any year!
Integrate[((2^Sqrt[x])/Sqrt[x]),x]=(2^(Sqrt[x]+1))/Ln[2]+C
too complicated solution. Just doing subtitution u=sqrt(x) and solution is lot easier ;)