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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = 2

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{1}^{4} (4 x^{2} + x) d x

\int \frac{1}{1 - 5 t} d t

\frac{d}{d x} (sin (x) + cos (2 x))

\int_{0}^{2} ∣ x - 1 ∣ d x

\int_{0}^{\infty} \frac{5}{x ^{2} + 4} d x

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