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

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

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

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

diverges

Solution steps

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{\partial}{\partial x} (r (θ, x) cos (θ))

x \to 3 - lim (\frac{3}{x ^{2} - 9})

\int_{1}^{3} \frac{x ^{2} + 5}{4 x - x ^{2}} d x

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

\frac{d}{d x} (15 x^{4})

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