Integration By Parts?

Question

Integral from 1 to E= (2t^2)(ln(t))dt

Answer 1

uv - integral vdu

let u = ln t, so du = 1/t dt
let dv = 2t^2, so v = (2/3)t^3

(2/3)t^3 * ln(t) - integral ((2/3)t^3*(1/t)) dt

the firts term is simplified as much as it can be, and now the second term is a simple integral to solve (just cancel out a t to simplify it and then solve it)

Answer 2

Formula for integrating by parts is:-
I = ∫ u .(dv/dt).dt = u.v - ∫ v.(du/dt).dt
Let u = ln t and dv/dt = 2.t²
du/dt = 1/t and v = 2t³ / 3
I = (ln t) (2t³/3) - ∫ (2t³/3).(1 / t).dt
I = (ln t).(2t³/3) - ∫ (2t² / 3). dt
I = (ln t).(2t³/3) - 2t³ / 9 between lims. of 1 to e
I = (ln e).(2e³/3) - 2.e³/9 - (- 2/9)
I = (ln e).(2e³/3) - 2.e³/9 + 2/9
I = 2e³/3 - 2.e³/9 + 2/9
I = 4e³/9 + 2/9
I = (2/9).(2e³ + 1)
See what you think of this----a bit tricky?