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

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

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

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

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

Decimal

1.31326 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{e - 1}^{e^{2} - 1}

Compute the boundaries:

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

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

\int_{0}^{2} x^{3} + 3 d x

\int_{1}^{3} x^{3} + 1 d x

\int_{0}^{2} f (3 x) d x

\int_{0}^{1} (2 - 3 x)^{4} d x

\int_{- \infty}^{\infty} e^{- j wt} d t

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