What is the integral from 1 to 10 of x/(x^2-4) ?

The integral from 1 to 10 of x/(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 1 to 10 of x/(x^2-4)

\int_{1}^{10} \frac{x}{x ^{2} - 4} d x

diverges

Solution steps

\int_{1}^{10} \frac{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_{1}^{2} \frac{x}{x ^{2} - 4} d x + \int_{2}^{10} \frac{x}{x ^{2} - 4} d x

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

g (x) = 2^{x^{2} + 1 x}

\frac{\partial}{\partial x} (x^{2} tan (4 x^{4} y))

\frac{d}{d x} (x^{2} sin^{3} (5 x))

\int \frac{1}{x ^{2} x ^{2} - 64} d x

\int \frac{1}{t ^{2} - 8 t + 25} d t

What is the integral from 1 to 10 of x/(x^2-4) ?
The integral from 1 to 10 of x/(x^2-4) is diverges