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

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

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 (1/(x^2)-1/x)

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

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

Decimal

- 0.19314 \dots

Solution steps

\int_{1}^{2} \frac{1}{x ^{2}} - \frac{1}{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{1}^{2} \frac{1}{x ^{2}} d x = \frac{1}{2}

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

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

\int_{1}^{4} π (\frac{2}{y})^{2} d y

\int_{0}^{1} 1 + r^{2} r d r

\int_{- 1}^{4} (- x^{2} + 6 x + 8 - x^{2}) d x

\int_{0}^{1} x^{- \frac{3}{2}} d x

\int_{0}^{y} sin (x + y) d x

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