What is the integral from 2 to 3 of 1/(x^2) ?

The integral from 2 to 3 of 1/(x^2) is 1/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 2 to 3 of 1/(x^2)

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

\frac{1}{6}

Decimal Notation

0.16666 \dots

Solution steps

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

Apply the Power Rule

: [- \frac{1}{x}]_{2}^{3}

= [- \frac{1}{x}]_{2}^{3}

Compute the boundaries:

\frac{1}{6}

= \frac{1}{6}

\int e^{\frac{x}{6} + 4} d x

\frac{d}{d x} (- 3 x^{6} ln (x))

\frac{d}{d x} ((4 - x^{\frac{2}{3}})^{\frac{3}{2}})

\frac{\partial}{\partial v} (\frac{1}{3} (v - 2 u))

\int (7 - x)^{49} d x

What is the integral from 2 to 3 of 1/(x^2) ?
The integral from 2 to 3 of 1/(x^2) is 1/6