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

The integral from 0 to 4 of (4x)/(x^2-4) 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 (4x)/(x^2-4)

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

diverges

Solution steps

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

\int_{0}^{2} \frac{4 x}{x ^{2} - 4} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{- 2}^{2} t^{3} (1 + t^{4})^{3} d t

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

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

\int_{4}^{3} sin (t) d t

\int_{0}^{1} 3 (1 - y)^{2} d y

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