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

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

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

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

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{1} \frac{1}{1 + u} d u

= \int_{0}^{1} \frac{1}{1 + u} d u

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= [ln ∣ v ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= ln (2)

\int_{1}^{2} x^{\frac{1}{8}} ln (x) d x

\int_{- 1}^{1} x - 1 d x

\int_{0}^{1 - y} 1 d z

\int_{0}^{1} x (1 - x)^{9} d x

\int_{- x}^{x} x^{2} d x

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