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

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

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

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

Decimal Notation

- 1.07398 \dots

Solution steps

\int_{1}^{2} \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}^{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}

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

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

= - 7 [- \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}

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

Compute the boundaries:

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

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

\int \frac{- 7}{cos ( x ) - 1} d x

\int tan^{3} (2 x) sec^{4} (2 x) d x

\int_{x}^{x + 2} (4 t + 1) d t

\frac{\partial}{\partial x} (2 cos (x y))

y = 2 x^{3} - 5 x, (1, - 3)

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