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

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}^{1} 2 t 4 t^{2} + 1 d t

\int_{1}^{8} (3 x + 2) d x

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

\int_{- 2}^{2} 2 - (x^{2} - 2) d x

\int_{0}^{t} sin (u) cos (t - u) d u

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