What is the integral from-2 to 2 of 1/(x+1) ?

The integral from-2 to 2 of 1/(x+1) is diverges

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-2 to 2 of 1/(x+1)

\int_{- 2}^{2} \frac{1}{x + 1} d x

diverges

Solution steps

\int_{- 2}^{2} \frac{1}{x + 1} d x

Find undefined (singularity) points:

x = - 1

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_{- 2}^{- 1} \frac{1}{x + 1} d x + \int_{- 1}^{2} \frac{1}{x + 1} d x

\int_{- 2}^{- 1} \frac{1}{x + 1} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{1} 2 t 1 + 4 t^{2} d t

\int_{- 1}^{1} \frac{x}{x ^{2} + x + 1} d x

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

\int_{0}^{π} (2 sin (x)) d x

\int_{1}^{e} ln^{3} (x) d x

What is the integral from-2 to 2 of 1/(x+1) ?
The integral from-2 to 2 of 1/(x+1) is diverges