What is the integral from 0 to 2pi of 1/x ?

The integral from 0 to 2pi of 1/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 2pi of 1/x

\int_{0}^{2 π} \frac{1}{x} d x

diverges

Solution steps

\int_{0}^{2 π} \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= [ln ∣ x ∣]_{0}^{2 π}

Compute the boundaries:

diverges

= diverges

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

\int_{0}^{\frac{π}{6}} 3 sec (θ) tan (θ) d θ

\int_{2}^{6} (6 x + 2)^{2} d x

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

\int_{1}^{3} (3 x^{2} - 2 x) d x

What is the integral from 0 to 2pi of 1/x ?
The integral from 0 to 2pi of 1/x is diverges