What is the integral from 1 to 3 of (e^x)/(e^x-1) ?

The integral from 1 to 3 of (e^x)/(e^x-1) is ln(e^3-1)-ln(e-1)

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integral from 1 to 3 of (e^x)/(e^x-1)

\int_{1}^{3} \frac{e ^{x}}{e ^{x} - 1} d x

ln (e^{3} - 1) - ln (e - 1)

Decimal Notation

2.40760 \dots

Solution steps

\int_{1}^{3} \frac{e ^{x}}{e ^{x} - 1} d x

Apply u-substitution

: \int_{e - 1}^{e^{3} - 1} \frac{1}{u} d u

= \int_{e - 1}^{e^{3} - 1} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{e - 1}^{e^{3} - 1}

Compute the boundaries:

ln (e^{3} - 1) - ln (e - 1)

= ln (e^{3} - 1) - ln (e - 1)

\int 3 - t e^{- t} d t

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\int 7 x e^{3 x} d x

\int 484 - x^{2} d x

\frac{\partial}{\partial x} (x e^{x^{2} + 2 y^{2}})

What is the integral from 1 to 3 of (e^x)/(e^x-1) ?
The integral from 1 to 3 of (e^x)/(e^x-1) is ln(e^3-1)-ln(e-1)