What is the integral from e to 6 of 1/(xln(x)) ?

The integral from e to 6 of 1/(xln(x)) is ln(ln(6))

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integral from e to 6 of 1/(xln(x))

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

ln (ln (6))

Solution steps

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

Apply u-substitution

: \int_{1}^{l n (6)} \frac{1}{u} d u

= \int_{1}^{l n (6)} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{l n (6)}

Compute the boundaries:

ln (ln (6))

= ln (ln (6))

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\int_{0}^{1} \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x

\int_{0}^{x} sinh^{2} (t) d t

\int_{0}^{t} e^{- a x} d x

What is the integral from e to 6 of 1/(xln(x)) ?
The integral from e to 6 of 1/(xln(x)) is ln(ln(6))