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Integrals- Log/Ln

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Can you check this please

i'm not sure why it's wrong .__.

1) indefinite integral of 1/t *ln(t) dt
the t*ln(t) is altogether on the bottom
i got ln(ln(x)) + C

2) find the indefinite integral of 1/(sqrt(t)*[1-2*sqrt(t)]dt
sqrt(x) is on the bottom.
i got -ln(1-2*sqrt(t))+C

can you tell me why mine's answers are wrong? thnx

  • Integrals- Log/Ln -

    1. u = ln(t)
    du = 1/t dt
    u du
    u^2/2 + c

    (ln(2))^2/2 + c

    2. 1/t^2 (1- 2t^2)dt
    Then multiply 1/t^2
    (1/t^2 -2)dt
    -1/t -2t + c

  • Integrals- Log/Ln -

    Typo
    (ln(t))^2/2 + c

  • Integrals- Log/Ln -

    I'm still getting it wrong, so I'm not sure what's with that.. Thank you for your help though!

  • Integrals- Log/Ln -

    #1 is correct: ln(lnx))+C

    #2: If you mean
    ∫ 1/√t * (1-2√t) dt
    Let u = 1-2√t
    du = -1/√t dt
    and you have
    ∫-u du
    = -1/2 u^2 + C
    = -1/2 (1-2√t)^2 + C
    = -1/2 (1 - 4√t + 4t) + C
    = -2 + 2√t - 2t + C
    you can absorb the -2 into C, leaving
    2√t - 2t + C

  • Integrals- Log/Ln -

    Or, if you mean
    ∫ 1/(√t * (1-2√t)) dt
    Again let u=1-2√t
    du = -1/√t dt
    and we have

    ∫ -1/u du
    = -lnu + C
    = -ln(1-2√t) + C

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