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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 1 to 2 of x^{-2}ln(x)

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

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

Decimal

0.15342 \dots

Solution steps

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

Apply Integration By Parts

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

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

= [- \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}

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

Compute the boundaries:

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

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

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