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

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

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

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

\frac{1}{2}

Decimal

0.5

Solution steps

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

Apply u-substitution

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

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

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

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

Take the constant out:

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

= - (- \int_{0}^{1} u d u)

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

\frac{\partial}{\partial b} (m x + b)

x \to 0 lim (e^{\frac{4}{x ^{2}}})

f (x) = 3 x^{4} (x - 3)^{2}

\frac{d}{d x} (- 3 sin (x) + cos (x))

y = 3^{- x^{2}}

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