What is the integral from 0 to infinity of 1/(4x+3) ?

The integral from 0 to infinity of 1/(4x+3) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to infinity of 1/(4x+3)

\int_{0}^{\infty} \frac{1}{4 x + 3} d x

diverges

Solution steps

\int_{0}^{\infty} \frac{1}{4 x + 3} d x

Compute the indefinite integral:

\int \frac{1}{4 x + 3} d x = \frac{1}{4} ln ∣ 4 x + 3 ∣ + C

Compute the boundaries:

\int_{0}^{\infty} \frac{1}{4 x + 3} d x =

diverges

= diverges

\int_{4}^{16} 2 x - 1 d x

\int_{0}^{x} x^{3} y^{2} d y

\int_{- 2}^{0} 2 x + 4 d x

\int_{\frac{1}{4}}^{\frac{1}{2}} \frac{2}{5} (x + 2) d x

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

What is the integral from 0 to infinity of 1/(4x+3) ?
The integral from 0 to infinity of 1/(4x+3) is diverges