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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

ln (3)

Decimal Notation

1.09861 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{3}

Compute the boundaries:

ln (3)

= ln (3)

\int_{0}^{6} 9 - (x - 3)^{2} d x

\int_{- 4}^{4} (x^{2} - 2) d x

\int_{0}^{π} x (π - x) sin (n x) d x

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

\int_{0}^{4} ∣ 2 x - 5 ∣ d x

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