Calculus AB

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Find the indefinite Integral of:

(e^(2x)-e^(-2x))/(e^(2x)+e^(-2x))dx

Thank you very much for your help.

  • Calculus AB -

    ∫(e^(2x)-e^(-2x))/(e^(2x)+e^(-2x))dx
    substitute
    sinh(2x)=(e^(2x)-e^(-2x))/2
    cosh(2x)=(e^(2x)+e^(-2x))/2
    =∫sinh(2x)/cosh(2x)dx
    Substitute
    u=cosh(2x)
    du=2sinh(2x)dx
    (1/2)du = sinh(2x)dx
    to get
    =(1/2)∫du/u
    =(1/2)log(u)+C
    =(1/2)log(cosh(2x))+C
    =(1/2)log((e^(2x)+e^(-2x))/2)+C

    Alternately, you could also substitute u=e^(2x)+e^(-2x)
    and do integration manually, with the same results.

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