What is the integral from 1 to 2 of 1/(x(1+ln(x))) ?

The integral from 1 to 2 of 1/(x(1+ln(x))) is ln(ln(2)+1)

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integral from 1 to 2 of 1/(x(1+ln(x)))

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

ln (ln (2) + 1)

Decimal Notation

0.52658 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{1 + l n (2)} \frac{1}{u} d u

= \int_{1}^{1 + l n (2)} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{1 + l n (2)}

Compute the boundaries:

ln (ln (2) + 1)

= ln (ln (2) + 1)

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