What is the integral from-1 to 1 of x^{-1} ?

The integral from-1 to 1 of x^{-1} is diverges

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integral from-1 to 1 of x^{-1}

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

diverges

Solution steps

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{0}^{π} x^{4} cos (2 x^{5}) d x

\int_{1}^{2} y^{2} d y

\int_{3}^{6} 2 d x

\int_{- 5}^{7} (2 x - 1) d x

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

What is the integral from-1 to 1 of x^{-1} ?
The integral from-1 to 1 of x^{-1} is diverges