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

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

Popular >

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

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

ln ∣ e^{x} + 1 ∣ - x + C

Solution steps

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

Take the constant out:

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

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

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

= - \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

= - (- \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

= - (- ln ∣ e^{x} + 1 ∣ + x)

Simplify

= ln ∣ e^{x} + 1 ∣ - x

Add a constant to the solution

= ln ∣ e^{x} + 1 ∣ - x + C

\int 60 (1 + x^{2}) (7 + 15 x + 5 x^{3})^{10} d x

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

\int \frac{e ^{3 x}}{( e ^{3 x} - 2 ) ^{6}} d x

\int \frac{( 7 x - 1 )}{( x - 1 ) ( x - 2 ) ( x - 3 )} d x

\int (\frac{1 + x}{x}) d x

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