What is the integral from 1 to e of ln(\sqrt[x]{x}) ?

The integral from 1 to e of ln(\sqrt[x]{x}) is 1/2

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 of ln(\sqrt[x]{x})

\frac{1}{2}

Decimal Notation

0.5

Solution steps

Simplify

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

Apply u-substitution

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

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

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{0}^{1}

= [\frac{u ^{2}}{2}]_{0}^{1}

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

\int_{0}^{\frac{π}{4}} sin (2 x + \frac{π}{2}) d x

\int_{0}^{\frac{π}{2}} x^{2} \cdot cos (x) d x

\int_{0}^{1} x^{2} \cdot 1 - x^{2} d x

\int_{0}^{5} (4 - x^{2}) d x

\int_{e}^{6} \frac{1}{x ln ( x )} d x

What is the integral from 1 to e of ln(\sqrt[x]{x}) ?
The integral from 1 to e of ln(\sqrt[x]{x}) is 1/2