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

y = \frac{x ^{2} + 5}{6 x - x ^{2}}, (1, 5)

\int - 7 lo g_{2} (x) d x

\int e^{x} (x + 1)^{2} d x

y = x^{2} - 4 x, y = 0

23 x^{3} + x^{2} - 12 x + 17

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