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

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

\int_{1}^{\infty} \frac{x}{2 x + 1} d x

diverges

Solution steps

\int_{1}^{\infty} \frac{x}{2 x + 1} d x

Compute the indefinite integral:

\int \frac{x}{2 x + 1} d x = \frac{1}{4} (2 x + 1 - ln ∣ 2 x + 1 ∣) + C

Compute the boundaries:

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

diverges

= diverges

\int_{0}^{5} \frac{x ^{2}}{25} d x

\int_{0}^{\infty} e^{- p x} d x

\int_{0}^{1} (x^{2} - 1) e^{x} d x

\int_{0}^{21} x (x + 4)^{\frac{1}{2}} d x

\int_{3}^{6} \frac{9}{x} d x

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