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

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

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

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

diverges

Solution steps

\int_{1}^{- 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}^{1} \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}^{1} \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_{0}^{3} (3 t - 1)^{58} d t

\int_{0}^{t} \frac{1}{x ^{2}} d x

\int_{1}^{4} 21 x^{\frac{5}{2}} d x

\int_{1}^{2.5} \frac{3}{2 x ^{2}} d x

\int_{0}^{1} t e^{- i wt} d t

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