What is the integral from 0 to e of 1/x ?

The integral from 0 to e of 1/x is diverges

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integral from 0 to e of 1/x

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

diverges

Solution steps

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

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= [ln ∣ x ∣]_{0}^{e}

Compute the boundaries:

diverges

= diverges

\int_{0}^{\frac{π}{3}} sin (2 x) d x

\frac{d}{d x} (ln (\frac{e ^{x} - 1}{e ^{x + 1}}))

f (x) = 19 x

\int e^{3 θ} sin (4 θ) d θ

\int_{0}^{2} (x - 7)^{2} d x

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