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

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

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

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

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

Decimal Notation

4.30548 \dots

Solution steps

\int_{1}^{8} 7 \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

= 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)

\frac{s - 2}{s ^{2} - 7 s + 6}

x^{- 6} + e^{2 x}

\frac{d y}{d x} + 2 x y = x

\int \frac{1}{24} x^{2} d x

\frac{\partial}{\partial x} (- x + 6 y)

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