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

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

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

\int_{- 1}^{1} (\frac{1}{x}) d x

diverges

Solution steps

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{2}^{8} \frac{x}{x + 2 x} d x

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

\int_{0}^{\frac{π}{2}} 7 sin (t) - 2 cos (t) d t

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

\int_{0}^{100} 1

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