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

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

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

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

diverges

Solution steps

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{1} \frac{1}{x ^{2} - 8 x + 15} d x

\int_{0}^{4} π (4 x)^{2} d x

\int_{0}^{1} (3 x^{3} - x^{2} + 2 x - 4) d x

\int_{0}^{1} \frac{32}{9} (x y)^{\frac{1}{3}} d y

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