What is the integral from 1 to infinity of 3/x ?

The integral from 1 to infinity of 3/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 3/x

\int_{1}^{\infty} \frac{3}{x} d x

diverges

Solution steps

\int_{1}^{\infty} \frac{3}{x} d x

Compute the indefinite integral:

\int \frac{3}{x} d x = 3 ln ∣ x ∣ + C

Compute the boundaries:

\int_{1}^{\infty} \frac{3}{x} d x =

diverges

= diverges

\int_{1}^{2} \frac{1}{( 1 - x ) ^{2}} d x

\int_{0}^{\frac{π}{3}} 6 sin (\frac{x}{2}) d x

\int_{0}^{x} cos (θ) d θ

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

\int_{0}^{2} t^{2} e^{- t^{3}} d t

What is the integral from 1 to infinity of 3/x ?
The integral from 1 to infinity of 3/x is diverges