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

The integral from 1 to e of 1/(x(2+ln(x))) is ln(3)-ln(2)

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

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

ln (3) - ln (2)

Decimal

0.40546 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{3} \frac{1}{u} d u

= \int_{2}^{3} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2}^{3}

Compute the boundaries:

ln (3) - ln (2)

= ln (3) - ln (2)

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

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

\int_{2}^{3} ∣ 1 - x ∣ d x

\int_{0}^{\infty} 1 e^{- x} d x

\int_{0}^{2 π} - cos (x) sin^{2} (x) d x

What is the integral from 1 to e of 1/(x(2+ln(x))) ?
The integral from 1 to e of 1/(x(2+ln(x))) is ln(3)-ln(2)