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

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

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

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

- \frac{2}{e} + 1

Decimal Notation

0.26424 \dots

Solution steps

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

Apply Integration By Parts

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

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

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

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

x \frac{1}{x} = 1

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

Compute the boundaries:

- \frac{2}{e} + 1

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

f (x) = x^{8}

\frac{d}{d x} (e^{- x} + x^{3} - 6)

\int 2 \cdot x d x

x \to 0 lim (\frac{sin ( 4 x )}{4})

y = \frac{2}{x - 2}

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