What is the integral from e to infinity of 1/(ln(x)) ?

The integral from e to infinity of 1/(ln(x)) is diverges

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integral from e to infinity of 1/(ln(x))

\int_{e}^{\infty} \frac{1}{ln ( x )} d x

diverges

Solution steps

\int_{e}^{\infty} \frac{1}{ln ( x )} d x

Compute the indefinite integral:

\int \frac{1}{ln ( x )} d x = Ei (ln (x)) + C

Compute the boundaries:

\int_{e}^{\infty} \frac{1}{ln ( x )} d x =

diverges

= diverges

\int_{- a}^{a} (2 sin (2 x)) d x

\int_{e}^{6} (\frac{1}{x ln ( x )}) d x

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

\int_{- 1}^{1} (∣ x ∣ - x^{2}) d x

\int_{0}^{1} (t + 1) e^{- t} d t

What is the integral from e to infinity of 1/(ln(x)) ?
The integral from e to infinity of 1/(ln(x)) is diverges