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

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

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

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

diverges

Solution steps

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int 2 e^{x} + 2 x - 1 d x

(x^{2} y^{3} - \frac{1}{1 + 9 x ^{2}}) d x + x^{3} y^{2} d y = 0

\frac{\partial}{\partial y} (ln (x^{4} + y) - z)

\frac{16}{( x ^{2} y ^{2} )}

\frac{d}{d y} (ln (x - 2 y))

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