Can We Integrate Lambert's W/x 😁

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[@Inspirator_AG112]

I have integrated W(1/x) visually before...

[@hemlataverma819]

An interesting one!

[@Nobodyman181]

SyberMath, if dx¹ is derivative, dx^(-1) is integral, can we found dx^(i)?

[@roberttelarket4934]

Can you integrate the Integral of SyberMath ds?

[@lazymello6778]

(s^2)(ybermath)/2 + c

[@roberttelarket4934]

@@lazymello6778: Ha ha ha ha!!! Very very very good!!!
You will receive the Cole prize!!!

[@lazymello6778]

@@roberttelarket4934 yay 🥳

[@Samiul_007]

Can you solve for y?
x = log(base b) of [ (W(y' /lnb)) /lna ]
let me know your answer 😁

[@maxvangulik1988]

u=W(x)
ue^u=x
(u+1)e^u•du=dx
I=int(u/(ue^u)•(u+1)e^u)du
I=int(u+1)du
I=(u+1)^2/2+C
I=(W(x)+1)^2/2+C

[@scottleung9587]

Cool!

[@goldfist8711]

nice

[@thexavier666]

In such problems, do we need to state/prove that W(x) is continuous at all points since we are integrating/differentiating?

[@galus_anonimus]

I wrote a whole explanation but yt decided to delete it :)
in short:
for Riemann integration: yeah you might do it to prove the function is integrable.
for others like the Lebesgue integral: it's not a requirement but it is definitely sufficient
Although I think that if you are able to provide a closed form integral function then the original function is most definitely integrable. Because well, there is a function thats derivative is the function in question.
BTW the Lambert W function is continous because it's one of the two branches of a functions that is a product of two contionous functions and that is strictly decreasing/increasing in each domain seperated by the number -1

[@FisicTrapella]

👌

[@GiuseppeAriano]

W(x)/x= e^-(W(x)). By parts xe^-(W(x))+int[xe^-(W(x))W'(x)]dx=xe^-(W(x))+int[W(x) e^(W(x))e^-(W(x))W'(x)]dx= easily the answer
Bye

[@phill3986]

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