Find the indefinite integral

(integral sign) 3te^2tdt

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)

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

then multiply by 3

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