What is the integral from 0 to pi of (e^x-1)/(e^x-x) ?

The integral from 0 to pi of (e^x-1)/(e^x-x) is ln(-pi+e^pi)

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integral from 0 to pi of (e^x-1)/(e^x-x)

\int_{0}^{π} \frac{e ^{x} - 1}{e ^{x} - x} d x

ln (- π + e^{π})

Decimal

2.99568 \dots

Solution steps

\int_{0}^{π} \frac{e ^{x} - 1}{e ^{x} - x} d x

Apply u-substitution

: \int_{1}^{e^{π} - π} \frac{1}{u} d u

= \int_{1}^{e^{π} - π} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{e^{π} - π}

Compute the boundaries:

ln (- π + e^{π})

= ln (- π + e^{π})

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What is the integral from 0 to pi of (e^x-1)/(e^x-x) ?
The integral from 0 to pi of (e^x-1)/(e^x-x) is ln(-pi+e^pi)