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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from e to infinity of 1/x

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

diverges

Solution steps

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

Compute the indefinite integral:

\int \frac{1}{x} d x = ln ∣ x ∣ + C

Compute the boundaries:

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

diverges

= diverges

\int_{0}^{3 t} \frac{sin ( u )}{u} d u

\int_{0}^{\infty} x \cdot e^{x} d x

\int_{0}^{π} x sin (k x) d x

\int_{0}^{y} \frac{2 x ^{2}}{y ^{2}} d x

\int_{\frac{1}{6}}^{e} ∣ ln (x) ∣ d x

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