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

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

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integral from-1 to 2 of 1/x

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

diverges

Solution steps

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

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\int_{0}^{\frac{π}{4}} (sec (x))^{2} d x

\int_{- π}^{0} cos (n x) d x

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