What is the integral of 2/(1+e^x) ?

The integral of 2/(1+e^x) is 2(-ln|1+e^x|+x)+C

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integral of 2/(1+e^x)

\int \frac{2}{1 + e ^{x}} d x

2 (- ln ∣ 1 + e^{x} ∣ + x) + C

Solution steps

\int \frac{2}{1 + e ^{x}} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{1 + e ^{x}} d x

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

= 2 \cdot \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

= 2 (- \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

= 2 (- ln ∣ 1 + e^{x} ∣ + x)

Add a constant to the solution

= 2 (- ln ∣ 1 + e^{x} ∣ + x) + C

\int_{- 1}^{3} 2 y - (y^{2} - 3) d y

\frac{d}{d x} (tan (x) \cdot sec (x))

n = 2 \sum \infty (- \frac{2}{3})^{n}

\int \frac{ln ( x )}{( x + 1 ) ^{2}} d x

\int_{1}^{2} (\frac{12}{x ^{2}} - 1) d x

What is the integral of 2/(1+e^x) ?
The integral of 2/(1+e^x) is 2(-ln|1+e^x|+x)+C