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

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

\frac{1}{2}

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}

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\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