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

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

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

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

- 2 ln ∣ 1 - e^{x} ∣ + x + C

Solution steps

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

\frac{1 + e ^{x}}{1 - e ^{x}} = \frac{2 e ^{x}}{1 - e ^{x}} + 1

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

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

h \to 0 lim (\frac{( 7 + h ) - 7}{h})

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

t^{2} (8 e^{2 t} - 6 sin (t))

x \to 0 lim ((1 + 4 x)^{\frac{4}{x}})

\frac{\partial}{\partial x} \frac{2 t}{( 1 - \frac{0.5 ( x )}{L} ) ^{2}}

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