how do you integrate e^x/1+e^x dx?

natasha-kelly_60b8bfcd11db8natasha-kelly_60b8bfcd11db8
in Science & Mathematics

please i really cant do it. i think you have you use integration by substitution.

Comments

  • neil-walsh_60b8bfce18315neil-walsh_60b8bfce18315

    I assume you meant ∫ e^(x) dx / [1 + e^(x)]

    Let u = 1 + e^(x), du = e^(x) dx

    ∫ e^(x) dx / [1 + e^(x)]

    = ∫ du / u

    = ln(u) + C

    = ln(1 + e^(x)) + C

  • rapidfirerapidfire

    Division always comes before addition unless there are brackets. Integration by substitution is not required for this integral.

    ∫ (e˟ / 1 + e˟) dx

    ∫ (e˟ + e˟) dx

    ∫ 2e˟ dx

    2 ∫ e˟ dx

    2e˟ + C

  • gagagoogoolologagagoogoololo

    You can rewrite it as e^x(1+e^x)^-1dx

    Then you can set u = 1 + e^x, giving you:

    du = e^xdx

    So you have the integral of (u^-1)du, which is ln(u), which is ln(1 + e^x) + C.

    The answer is ln(1 + e^x) + C.

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