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

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

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

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

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

Decimal

1.38083 \dots

Solution steps

\int_{1}^{2} 9 \cdot \frac{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{1}{x ( x + 1 )} d x

f (x) = x - 1, (10, 3)

\int_{0}^{3} 4 x + 5 d x

x = - 3, x = - 1, y = x^{3} + 8, y = 0

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

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