What is the integral from e^1 to e^2 of 1/(xln(x)) ?

The integral from e^1 to e^2 of 1/(xln(x)) is ln(2)

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integral from e^1 to e^2 of 1/(xln(x))

\int_{e^{1}}^{e^{2}} \frac{1}{x ln ( x )} d x

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

\int_{e^{1}}^{e^{2}} \frac{1}{x ln ( x )} d x

Apply u-substitution

: \int_{1}^{2} \frac{1}{u} d u

= \int_{1}^{2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= ln (2)

\int_{0}^{3} - 9 - x^{2} d x

\int_{- 1}^{2} 5^{3 x} d x

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

\int_{0}^{3} r e^{- r^{2}} d r

\int_{0}^{t} e^{- 4.2804 x} d x

What is the integral from e^1 to e^2 of 1/(xln(x)) ?
The integral from e^1 to e^2 of 1/(xln(x)) is ln(2)