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

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

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integral of (e^x-1)/x

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

Ei (x) - ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

\int \frac{e ^{x}}{x} d x = Ei (x)

\int \frac{1}{x} d x = ln ∣ x ∣

= Ei (x) - ln ∣ x ∣

Add a constant to the solution

= Ei (x) - ln ∣ x ∣ + C

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