What is the integral from 3 to infinity of 1/(2x-1) ?

The integral from 3 to infinity of 1/(2x-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 3 to infinity of 1/(2x-1)

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

diverges

Solution steps

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

Compute the indefinite integral:

\int \frac{1}{2 x - 1} d x = \frac{1}{2} ln ∣ 2 x - 1 ∣ + C

Compute the boundaries:

\int_{3}^{\infty} \frac{1}{2 x - 1} d x =

diverges

= diverges

\int_{- 3}^{4} x^{2} - 4 d x

\int_{1}^{\infty} 3 x^{- 2} d x

\int_{0}^{π} 7 x sin (x) d x

\int_{7}^{- 5} g (x) d x

\int_{0}^{8} 2 x e^{- 0.2 x} d x

What is the integral from 3 to infinity of 1/(2x-1) ?
The integral from 3 to infinity of 1/(2x-1) is diverges