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

The integral from 0 to u 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 u of 1/x

\int_{0}^{u} \frac{1}{x} d x

diverges

Solution steps

\int_{0}^{u} \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= [ln ∣ x ∣]_{0}^{u}

Compute the boundaries:

diverges

= diverges

\int_{- 4}^{4} - 5 d x

\int_{2}^{4} 4 x^{2} - 5 x + 8 d x

\int_{0}^{3} \frac{x e ^{x}}{( x + 1 ) ^{2}} d x

\int_{- 2}^{2} (4 - y^{2}) - (y^{2} - 4) d y

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

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