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

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

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

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

diverges

Solution steps

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Find undefined (singularity) points:

x = 0

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{- \infty}^{- 2} \frac{1}{x} d x

\int_{0}^{1} (x^{2} + 8) d x

\int_{- 1}^{3} (- x^{2} + 5 x) - (3 x - 3) d x

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

\int_{0}^{π} 4 sin (t) d t

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