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

The integral from 0 to 1 of (e^x)/(e^x-1) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

diverges

Solution steps

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

Apply u-substitution

: \int_{0}^{e - 1} \frac{1}{u} d u

= \int_{0}^{e - 1} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{0}^{e - 1}

Compute the boundaries:

diverges

= diverges

\int_{0}^{\frac{π}{4}} cos^{2} (3 x) d x

\int_{0}^{t a n (x)} e^{- t^{2}} d t

\int_{3}^{6} \frac{1}{x ^{2} - 3 x + 2} d x

\int_{3}^{5} 2 x d x

\int_{π}^{2 π} x sin (x) d x

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