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

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

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integral from 1 to e^3 of (1-ln(x))/x

\int_{1}^{e^{3}} \frac{1 - ln ( x )}{x} d x

- \frac{3}{2}

Decimal

- 1.5

Solution steps

\int_{1}^{e^{3}} \frac{1 - ln ( x )}{x} d x

Apply u-substitution

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

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{- 2}^{1} u d u)

Apply the Power Rule

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

= - (- [\frac{u ^{2}}{2}]_{- 2}^{1})

Simplify

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

Compute the boundaries:

- \frac{3}{2}

= - \frac{3}{2}

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