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

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

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

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

- \frac{- ln ( 2 ) - 1}{2} - 1

Decimal

- 0.15342 \dots

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

Compute the boundaries:

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

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

Simplify

= - \frac{- ln ( 2 ) - 1}{2} - 1

\frac{\partial}{\partial x} (2 x^{2} + y^{2})

n \to \infty lim (1.0 4^{n})

\frac{d}{d x} (- x^{3} + 6 x^{2} - 9 x)

\int \frac{- 5}{x ^{2} + 16} d x

\int \frac{2 e ^{- \frac{6}{x ^{4}}}}{x ^{5}} d x

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