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

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

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

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

\frac{1}{3}

Decimal

0.33333 \dots

Solution steps

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

Apply u-substitution

: \int_{0}^{1} u^{2} d u

= \int_{0}^{1} u^{2} d u

Apply the Power Rule

: [\frac{u ^{3}}{3}]_{0}^{1}

= [\frac{u ^{3}}{3}]_{0}^{1}

Compute the boundaries:

\frac{1}{3}

= \frac{1}{3}

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