What is the integral from 1 to 8 of 3(ln(x))/(x^2) ?

The integral from 1 to 8 of 3(ln(x))/(x^2) is 3((-3ln(2)-1)/8+1)

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integral from 1 to 8 of 3(ln(x))/(x^2)

\int_{1}^{8} 3 \frac{ln ( x )}{x ^{2}} d x

3 (\frac{- 3 ln ( 2 ) - 1}{8} + 1)

Decimal Notation

1.84520 \dots

Solution steps

\int_{1}^{8} 3 \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

= 3 \cdot \int_{1}^{8} \frac{ln ( x )}{x ^{2}} d x

Apply Integration By Parts

: [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{8}

= 3 [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{8}

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

= 3 [- \frac{ln ( x )}{x} - \frac{1}{x}]_{1}^{8}

Since the denominators are equal, combine the fractions:

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

= 3 [\frac{- ln ( x ) - 1}{x}]_{1}^{8}

Compute the boundaries:

\frac{- 3 ln ( 2 ) - 1}{8} + 1

= 3 (\frac{- 3 ln ( 2 ) - 1}{8} + 1)

(x^{2} + x y - y^{2}) d x + (x y) d y = 0

ln (x + 2), 3

x \to 0 lim (x \cdot ln (x^{2}))

\frac{d}{d t} (e^{t a n (t)})

\frac{d ^{4}}{d x ^{4}} (e^{x^{2}})

What is the integral from 1 to 8 of 3(ln(x))/(x^2) ?
The integral from 1 to 8 of 3(ln(x))/(x^2) is 3((-3ln(2)-1)/8+1)