Calculus
posted by Barbara .
Find the indefinite integral
(integral sign) 3te^2tdt

use a calculator, that's what calculus is all about. duh!

Excuse you that DUH is not necessary and if I could use the calculator I would have used it but my teacher did not teach me to how to use the calculator for this so please mind your business

do t e^(2t) dt
and multiply by 3 later
by parts
u = t
du = dt
dv = e^(2t) dt
v = (1/2) e^(2t)
u v = (1/2) t e^(2t)
v du = (1/2)e^(2t) dt
integral v du = e^(2t)
u v  integral v du = (1/2) t e^(2t) e^(2t) 
then multiply by 3

integral(u dv) = uv  integral(v du)
let u = 3t
du/dt = 3 , so du = 3 dt
let dv = e^(2t)dt
v = (1/2)e^(2t)
so integral(3t(e^(2t)))dt = 3t(1/2)e^(2t)  integral((1/2)e^(2t)(3 dt)
= (3/2)te^(2t)  (3/4)e^(2t)
I don't know how much further you want to simplify this, but I differentiated my last answer and it works 
Let 3t = u and e^2t dt = dv
du = 3 dt v = (1/2) e^2t
The integral is
uv  INTEGRAL v du
= (3/2)t e^2t  INTEGRAL (3/2)e^2t
= (3/2)t e^2t  (3/4)e^2t 
Thank You Damon, Reiny and Drwls for your help

but why does v=(1/2)e^2t

all three of us had chosen
let dv = e^(2t)dt
or dv/dt = e^(2t)
wouldn't you have to integrate that to get v ?
v = (1/2)e^(2t)
I hope you recognized that we used a method called integration by parts
in choosing the "u" and "dv"
let u be something that you can differentiate, and
let dv be the part that you can integrate,
then hope for the best 
because d/dt of e^2t = 2 e^2t
so you need the (1/2) to get one of them instead of 2 
oooo okay Thank You

yeah you better say thank you and if i were to mind my own business then why did you post a question where everyone can see dufus?

Ummm because I wasn't looking for help from you and of course I'm going to say Thank You because i have manners unlike you you don't have any and you don't address people that's looking for help form you like that so Goodbye and Have a nice day