What is the integral from 1 to 3 of x/(4-x^2) ?

The integral from 1 to 3 of x/(4-x^2) is diverges

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integral from 1 to 3 of x/(4-x^2)

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

diverges

Solution steps

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{- 1}^{2} (x^{3} - 6 x) d x

\int_{0}^{2} (x^{3} - 4 x + 3) d x

\int_{0}^{1} x (x^{2} + 3)^{n} d x

\int_{- 2}^{0} \frac{1}{x ^{3}} d x

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

What is the integral from 1 to 3 of x/(4-x^2) ?
The integral from 1 to 3 of x/(4-x^2) is diverges