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

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

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

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

ln (- 1 + e) - ln (- 1 + e^{2}) + 1

Decimal

- 0.31326 \dots

Solution steps

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

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

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

\int_{1}^{2} \frac{e ^{x}}{1 - e ^{x}} d x = ln (- 1 + e) - ln (- 1 + e^{2})

\int_{1}^{2} 1 d x = 1

= ln (- 1 + e) - ln (- 1 + e^{2}) + 1

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