What is the integral from 1 to 8 of (5ln(x))/(x^2) ?

The integral from 1 to 8 of (5ln(x))/(x^2) is 5((-3ln(2)-1)/8+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 8 of (5ln(x))/(x^2)

\int_{1}^{8} \frac{5 ln ( x )}{x ^{2}} d x

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

Decimal Notation

3.07534 \dots

Solution steps

\int_{1}^{8} \frac{5 ln ( x )}{x ^{2}} d x

Take the constant out:

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

= 5 \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}

= 5 [- \frac{ln ( x )}{x} - \int - \frac{1}{x ^{2}} d x]_{1}^{8}

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

= 5 [- \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}

= 5 [\frac{- ln ( x ) - 1}{x}]_{1}^{8}

Compute the boundaries:

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

= 5 (\frac{- 3 ln ( 2 ) - 1}{8} + 1)

n = 0 \sum \infty n - 2

y^{^{''}} + 2 y^{^{'}} + 3 y = 0, y^{^{'}} (0) = 2, y (0) = 1

y^{^{'}} + 7 y = e^{3 x}

\int_{0}^{\infty} x e^{4 x} d x

2^{\frac{1}{x}}

What is the integral from 1 to 8 of (5ln(x))/(x^2) ?
The integral from 1 to 8 of (5ln(x))/(x^2) is 5((-3ln(2)-1)/8+1)