What is the integral from 0 to 1 of 1/u ?

The integral from 0 to 1 of 1/u is diverges

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integral from 0 to 1 of 1/u

\int_{0}^{1} \frac{1}{u} d u

diverges

Solution steps

\int_{0}^{1} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{0}^{1}

Compute the boundaries:

diverges

= diverges

\int_{- 2}^{- 1} ∣ x ∣ d x

\int_{0}^{1} sin (w x) d x

\int_{\frac{π}{6}}^{\frac{π}{4}} csc^{2} (x) d x

\int_{0}^{π} x \cdot sin (3 x) d x

\int_{1}^{2} x e^{4 x^{2}} d x

What is the integral from 0 to 1 of 1/u ?
The integral from 0 to 1 of 1/u is diverges