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)

f (x) = x^{2} - \frac{1}{x}, at x = 2

\frac{d}{d y} (e^{\frac{y}{x}})

y^{^{'}} = \frac{( e ^{x} )}{y}

\int (e^{3 x} - 1)^{3} d x

- ln (sin (t))

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)