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

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

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

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

- 2 (- \frac{2}{e} + 1)

Decimal

- 0.52848 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

= - 2 [\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}

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

x \frac{1}{x} = 1

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

Compute the boundaries:

- \frac{2}{e} + 1

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

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