What is the integral from e^{-1} to e^1 of |ln(x)| ?

The integral from e^{-1} to e^1 of |ln(x)| is -2/e+2

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integral from e^{-1} to e^1 of |ln(x)|

\int_{e^{- 1}}^{e^{1}} ∣ ln (x) ∣ d x

- \frac{2}{e} + 2

Solution steps

\int_{e^{- 1}}^{e^{1}} ∣ ln (x) ∣ d x

Apply u-substitution

: \int_{- 1}^{1} ∣ u ∣ e^{u} d u

= \int_{- 1}^{1} ∣ u ∣ e^{u} d u

Eliminate Absolutes

= \int_{- 1}^{0} (- u) e^{u} d u + \int_{0}^{1} u e^{u} d u

\int_{- 1}^{0} (- u) e^{u} d u = 1 - \frac{2}{e}

\int_{0}^{1} u e^{u} d u = 1

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

Simplify

= - \frac{2}{e} + 2

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What is the integral from e^{-1} to e^1 of |ln(x)| ?
The integral from e^{-1} to e^1 of |ln(x)| is -2/e+2