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

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

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

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

x + \frac{1}{e ^{x}} + C

Solution steps

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

Expand

\frac{e ^{x} - 1}{e ^{x}} : 1 - \frac{1}{e ^{x}}

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

\int 1 d x = x

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

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

Simplify

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

Add a constant to the solution

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

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