What is the integral from-1 to 3 of 1/x ?

The integral from-1 to 3 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-1 to 3 of 1/x

\int_{- 1}^{3} \frac{1}{x} d x

diverges

Solution steps

\int_{- 1}^{3} \frac{1}{x} d x

Find undefined (singularity) points:

x = 0

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

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

\int_{- 1}^{0} \frac{1}{x} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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

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

\int_{0}^{2} 4 x (x^{2} + 3)^{3} d x

\int_{2}^{3} (x^{2} - 2 x - 15) d x

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

What is the integral from-1 to 3 of 1/x ?
The integral from-1 to 3 of 1/x is diverges