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

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

\int_{0}^{x} x + 2 y d y

\int_{1}^{2} (3 x^{2} - 2 x - 1) d x

\int_{1}^{2} (x - 2)^{5} d x

\int_{- 5}^{- 3} - x^{2} - 3 x d x

\int_{0}^{4} (3 x + 4) d x

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