What is the integral from 1 to 2 of (-9ln(x))/(x^2) ?

The integral from 1 to 2 of (-9ln(x))/(x^2) is -9((-ln(2)-1)/2+1)

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integral from 1 to 2 of (-9ln(x))/(x^2)

\int_{1}^{2} \frac{- 9 ln ( x )}{x ^{2}} d x

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

Solution steps

\int_{1}^{2} \frac{- 9 ln ( x )}{x ^{2}} d x

Take the constant out:

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

= - 9 \cdot \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}

= - 9 [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{2}

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

= - 9 [- \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}

= - 9 [\frac{- ln ( x ) - 1}{x}]_{1}^{2}

Compute the boundaries:

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

= - 9 (\frac{- ln ( 2 ) - 1}{2} + 1)

\int \frac{x + 2}{x - 3} d x

\int (\frac{5 x ^{2} + x - 4}{x ^{2}}) d x

y = - 2 t^{3}, (- 1, 2)

\frac{\partial}{\partial y} (\frac{3 x y}{x ^{4} + y ^{4}})

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

What is the integral from 1 to 2 of (-9ln(x))/(x^2) ?
The integral from 1 to 2 of (-9ln(x))/(x^2) is -9((-ln(2)-1)/2+1)