What is the integral from e^3 to e^5 of 1/(xln(x)) ?

The integral from e^3 to e^5 of 1/(xln(x)) is ln(5)-ln(3)

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integral from e^3 to e^5 of 1/(xln(x))

\int_{e^{3}}^{e^{5}} \frac{1}{x ln ( x )} d x

ln (5) - ln (3)

Solution steps

\int_{e^{3}}^{e^{5}} \frac{1}{x ln ( x )} d x

Apply u-substitution

: \int_{3}^{5} \frac{1}{u} d u

= \int_{3}^{5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{3}^{5}

Compute the boundaries:

ln (5) - ln (3)

= ln (5) - ln (3)

\int_{0}^{x} 2 x + 1 d x

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\int_{- 1}^{1} x ∣ 2 x - 1 ∣ d x

\int_{0}^{1} (x + 1) e^{2 x} d x

\int_{2}^{4} x^{3} + 4 x^{2} d x

What is the integral from e^3 to e^5 of 1/(xln(x)) ?
The integral from e^3 to e^5 of 1/(xln(x)) is ln(5)-ln(3)