What is the integral from 1 to infinity of e^{1/x} ?

The integral from 1 to infinity of e^{1/x} is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to infinity of e^{1/x}

\int_{1}^{\infty} e^{\frac{1}{x}} d x

diverges

Solution steps

\int_{1}^{\infty} e^{\frac{1}{x}} d x

Compute the indefinite integral:

\int e^{\frac{1}{x}} d x = e^{\frac{1}{x}} x - Ei (\frac{1}{x}) + C

Compute the boundaries:

\int_{1}^{\infty} e^{\frac{1}{x}} d x =

diverges

= diverges

\int_{2}^{4} 4 - x d x

\int_{0}^{2} \frac{x}{x ^{2} + 6 x + 8} d x

\int_{- 4}^{4} [(32 - x^{2}) - (x^{2})] d x

\int_{0}^{0.41} ln (1 + x^{2}) d x

\int_{4}^{5} \frac{1}{5 - x} d x

What is the integral from 1 to infinity of e^{1/x} ?
The integral from 1 to infinity of e^{1/x} is diverges