What is the integral from 1 to e of (ln(t))/t ?

The integral from 1 to e of (ln(t))/t is 1/2

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integral from 1 to e of (ln(t))/t

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

\frac{1}{2}

Solution steps

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

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}

n = 1 \sum \infty \frac{2 ^{n}}{3 ^{n}}

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\frac{d ^{2}}{d x ^{2}} (\frac{4}{5 t ^{6}})

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