What is the integral from 1 to 9 of 1/(x-5) ?

The integral from 1 to 9 of 1/(x-5) 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 9 of 1/(x-5)

\int_{1}^{9} \frac{1}{x - 5} d x

diverges

Solution steps

\int_{1}^{9} \frac{1}{x - 5} d x

Find undefined (singularity) points:

x = 5

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}^{5} \frac{1}{x - 5} d x + \int_{5}^{9} \frac{1}{x - 5} d x

\int_{1}^{5} \frac{1}{x - 5} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

x \to 0 lim (x cos (\frac{1}{x}))

\int x y e^{x + y} d x

\frac{d}{d x} (\frac{( a x + b y ( x ) )}{( c x + d y )})

(x^{2} - 2) (x^{3} - 3 x)

\int \frac{tan ( 3 x )}{cos ( 3 x )} d x

What is the integral from 1 to 9 of 1/(x-5) ?
The integral from 1 to 9 of 1/(x-5) is diverges