Indefinite Integrals

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find the indefinite integral of

(1/x)*(e^(-2log[ 3,x])) dx

where 3 is base of the log

i cannot even figure out where to start :( please help. thank you

  • Indefinite Integrals -

    log_3(x) = lnx/ln3

    = e^(-2lnx/ln3)
    = (e^(lnx))^(-2/ln3)
    = x^(-2/ln3)

    so you have
    ∫(1/x) x^(-2/ln3) dx
    = ∫x^(-2/ln3-1) dx
    = 1/(-2/ln3) x^(-2/ln3)
    -ln3/2 x^(-2/ln3)

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