What is the integral from 0 to infinity of 1/(7+x) ?

The integral from 0 to infinity of 1/(7+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 0 to infinity of 1/(7+x)

\int_{0}^{\infty} \frac{1}{7 + x} d x

diverges

Solution steps

\int_{0}^{\infty} \frac{1}{7 + x} d x

Compute the indefinite integral:

\int \frac{1}{7 + x} d x = ln ∣ 7 + x ∣ + C

Compute the boundaries:

\int_{0}^{\infty} \frac{1}{7 + x} d x =

diverges

= diverges

\int e^{2} x d x

(e^{x} + e^{- x}) y^{^{'}} = y^{2}

\int x^{- 4} + 2 x + 3 d x

x \to 0 lim (\frac{2 1 ^{x} - 1}{x})

\int \frac{1}{2 x ( 1 + x )} d x

What is the integral from 0 to infinity of 1/(7+x) ?
The integral from 0 to infinity of 1/(7+x) is diverges