What is the integral from 0 to 1 of x^{-1} ?

The integral from 0 to 1 of x^{-1} is diverges

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integral from 0 to 1 of x^{-1}

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

diverges

Solution steps

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

Use the common integral:

\int x^{- 1} d x = ln (∣ x ∣)

= [ln ∣ x ∣]_{0}^{1}

Compute the boundaries:

diverges

= diverges

y = 4 cot (3 - 2 x)

\int_{0}^{2} (4 x - 2 x^{2}) d x

x \to 2 lim (\frac{x}{( x - 2 )})

\int \frac{20}{x 1 + 2 x} d x

\int \frac{x}{x ^{2} + 2 x + 5} d x

What is the integral from 0 to 1 of x^{-1} ?
The integral from 0 to 1 of x^{-1} is diverges