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Evaluate the following definite Integral
∫x * ln(x) dx from 1 to e^(2)

I used the DI method which gave me
D I
+ ln(x) x
- 1/x (x^2)/2

= (ln(x)*x^(2))/2 - ∫ 1/x * x^(2)/2
= (ln(x)*x^(2))/2 - ∫ x/2
= (ln(x)*x^(2))/2 - x^(2)/4 + C

(ln(x)*x^(2))/2 - x^(2)/4| from 1 to e^(2)
= (ln(e^(2))*e^(2)^(2))/2 - e^(2)^(2)/4 - [ (ln(1)*1^(2))/2 - 1^(2)/4 ]
= (ln(e^(2))*e^(4)/2 - e^(4)/4 - [ (0*1)/2 - 1/4 ]
= (1*2)*e^(4)/2 - e^(4)/4 - [ 0 - 1/4 ]
= 2*e^(4)/2 - e^(4)/4+ 1/4

The teacher answer is
e^(2)/2 - e^(4)/4 + 1/4

Where did I go wrong? Thank you!

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  1. ∫x * ln(x)
    use integration by parts. That is just the product rule for derivatives, only in reverse. So, let
    u = lnx
    du = 1/x dx
    dv = x dx
    v = 1/2 x^2
    ∫u dv = uv - ∫v du
    ∫x lnx dx = 1/2 x^2 lnx - ∫1/2 x dx
    = 1/2 x^2 lnx - 1/4 x^2 = 1/4 x^2 (2lnx - 1)
    ∫[1,e^2] x lnx dx = (1/4 e^4 (2lne^2 - 1) - (1/4 (2ln1 -1))
    = 1/4 e^4 (3) - 1/4 (-1)
    = 1/4 (3e^4 + 1)

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    oobleck
  2. = 2*e^(4)/2 - e^(4)/4+ 1/4
    = 2e^(4)/2 - e^(4)/4 + 1/4
    = 2*2e^(4)/4 - e^(4)/4 + 1/4
    = 4e^(4)/4 - e^(4)/4 + 1/4
    = 3e^(4)/4 + 1/4
    = 1/4 [3e^(4) +1]

    Alright, thank you. I just couldnt see it somehow

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  3. But wait... is the teacher answer: e^(2)/2 - e^(4)/4 + 1/4, a mistake? like the e^(2)/2 part?

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