What is the integral of ((1-e^{ax})/(e^{ax)}) ?

The integral of ((1-e^{ax})/(e^{ax)}) is x/(e^{ax)}-e^{-ax}x-(e^{-ax})/a-x+C

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integral of ((1-e^{ax})/(e^{ax)})

\int (\frac{1 - e ^{a x}}{e ^{a x}}) d x

\frac{x}{e ^{a x}} - e^{- a x} x - \frac{e ^{- a x}}{a} - x + C

Solution steps

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

Expand

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

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

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

\int 1 d x = x

= \frac{x}{e ^{a x}} - e^{- a x} x - \frac{e ^{- a x}}{a} - x

Add a constant to the solution

= \frac{x}{e ^{a x}} - e^{- a x} x - \frac{e ^{- a x}}{a} - x + C

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