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

The integral from 1 to e of (ln(z))/(z^2) is -2/e+1

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

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

- \frac{2}{e} + 1

Decimal Notation

0.26424 \dots

Solution steps

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

Apply Integration By Parts

: [\frac{- ln ( z ) - z \cdot \int - \frac{1}{z ^{2}} d z}{z}]_{1}^{e}

= [\frac{- ln ( z ) - z \cdot \int - \frac{1}{z ^{2}} d z}{z}]_{1}^{e}

\int - \frac{1}{z ^{2}} d z = \frac{1}{z}

= [\frac{- ln ( z ) - z \frac{1}{z}}{z}]_{1}^{e}

z \frac{1}{z} = 1

= [\frac{- ln ( z ) - 1}{z}]_{1}^{e}

Compute the boundaries:

- \frac{2}{e} + 1

= - \frac{2}{e} + 1

\frac{d}{d x} (\frac{80}{1 + 0.32 x})

x \to 1 lim (e^{x - 1})

\frac{d}{d x} (- \frac{2}{2 x + 5})

f (x) = ln (4 x^{2} + 1)

x \to 1 - lim (\frac{2 x}{x ^{2} - 1})

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