What is the integral from 0 to 4 of 1/(x^2-6x+5) ?

The integral from 0 to 4 of 1/(x^2-6x+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 0 to 4 of 1/(x^2-6x+5)

\int_{0}^{4} \frac{1}{x ^{2} - 6 x + 5} d x

diverges

Solution steps

\int_{0}^{4} \frac{1}{x ^{2} - 6 x + 5} 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_{0}^{1} \frac{1}{x ^{2} - 6 x + 5} d x + \int_{1}^{4} \frac{1}{x ^{2} - 6 x + 5} d x

\int_{0}^{1} \frac{1}{x ^{2} - 6 x + 5} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

(e^{- 3 x})^{^{'}}

\frac{d}{d x} (ln (\frac{x + 1}{x ^{3} + 1}))

(- 16 x^{2} + 36)^{2}

\frac{d}{d x} (cos (3 x^{2} + x - 1))

\frac{d}{d x} (ln (x) + \frac{x - 1}{x})

What is the integral from 0 to 4 of 1/(x^2-6x+5) ?
The integral from 0 to 4 of 1/(x^2-6x+5) is diverges