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

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

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integral of 2/(e^x+1)

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

2 (- ln ∣ e^{x} + 1 ∣ + x) + C

Solution steps

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

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

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

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

\int 1 d x = x

= 2 (- ln ∣ e^{x} + 1 ∣ + x)

Add a constant to the solution

= 2 (- ln ∣ e^{x} + 1 ∣ + x) + C

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