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

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

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

integral

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

\frac{1}{e} ln ∣ x ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{e} \cdot \int \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= \frac{1}{e} ln ∣ x ∣

Add a constant to the solution

= \frac{1}{e} ln ∣ x ∣ + C