What is the integral of-1/(1+e^x) ?

The integral of-1/(1+e^x) is ln|1+e^x|-x+C

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integral of-1/(1+e^x)

\int - \frac{1}{1 + e ^{x}} d x

ln ∣ 1 + e^{x} ∣ - x + C

Solution steps

\int - \frac{1}{1 + e ^{x}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{1 + e ^{x}} d x

\frac{1}{1 + e ^{x}} = - \frac{e ^{x}}{1 + e ^{x}} + 1

= - \int - \frac{e ^{x}}{1 + e ^{x}} + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- \int \frac{e ^{x}}{1 + e ^{x}} d x + \int 1 d x)

\int \frac{e ^{x}}{1 + e ^{x}} d x = ln ∣ 1 + e^{x} ∣

\int 1 d x = x

= - (- ln ∣ 1 + e^{x} ∣ + x)

Simplify

= ln ∣ 1 + e^{x} ∣ - x

Add a constant to the solution

= ln ∣ 1 + e^{x} ∣ - x + C

\int e^{100 t} \cdot 60 d t

x \to 3 lim (\frac{4}{x - 3})

f (x) = 3 (x - 2)^{2}

x y^{^{'}} + y = x^{4} y^{4}, y (1) = \frac{1}{2}

y = e^{- m x}