What is the integral from-1 to-2 of (e^x)/(1-e^x) ?

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

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integral from-1 to-2 of (e^x)/(1-e^x)

\int_{- 1}^{- 2} \frac{e ^{x}}{1 - e ^{x}} d x

- ln (\frac{e ^{2} - 1}{e ^{2} - e})

Decimal Notation

- 0.31326 \dots

Solution steps

\int_{- 1}^{- 2} \frac{e ^{x}}{1 - e ^{x}} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 2}^{- 1} \frac{e ^{x}}{1 - e ^{x}} d x

Apply u-substitution

: \int_{1 - \frac{1}{e ^{2}}}^{1 - \frac{1}{e}} - \frac{1}{u} d u

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - (- (- [ln ∣ u ∣]_{1 - \frac{1}{e}}^{1 - \frac{1}{e ^{2}}}))

Simplify

= - [ln ∣ u ∣]_{1 - \frac{1}{e}}^{1 - \frac{1}{e ^{2}}}

Compute the boundaries:

ln (1 - \frac{1}{e ^{2}}) - ln (1 - \frac{1}{e})

= - (ln (1 - \frac{1}{e ^{2}}) - ln (1 - \frac{1}{e}))

Simplify

= - ln (\frac{e ^{2} - 1}{e ^{2} - e})

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