What is the integral from 1 to 8 of (7ln(x))/(x^2) ?

The integral from 1 to 8 of (7ln(x))/(x^2) is 7((-3ln(2)-1)/8+1)

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integral from 1 to 8 of (7ln(x))/(x^2)

\int_{1}^{8} \frac{7 ln ( x )}{x ^{2}} d x

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

Decimal

4.30548 \dots

Solution steps

\int_{1}^{8} \frac{7 ln ( x )}{x ^{2}} d x

Take the constant out:

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

= 7 \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}

= 7 [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{8}

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

= 7 [- \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}

= 7 [\frac{- ln ( x ) - 1}{x}]_{1}^{8}

Compute the boundaries:

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

= 7 (\frac{- 3 ln ( 2 ) - 1}{8} + 1)

x \to 2 lim (\frac{1}{x} - \frac{1}{x ^{2}})

\frac{π}{6}

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

\frac{d}{d x} (co tan (x))

\frac{d}{d x} (arccos (sin (x)))

What is the integral from 1 to 8 of (7ln(x))/(x^2) ?
The integral from 1 to 8 of (7ln(x))/(x^2) is 7((-3ln(2)-1)/8+1)