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

The integral from 1 to 2 of (1+x)^{-1} is ln(3)-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)^{-1}

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

ln (3) - ln (2)

Decimal

0.40546 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{3} \frac{1}{u} d u

= \int_{2}^{3} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2}^{3}

Compute the boundaries:

ln (3) - ln (2)

= ln (3) - ln (2)

\int_{0}^{4} \frac{2}{x + 1} d x

\int_{- 3}^{3} x^{3} + 4 x + 1 d x

\int_{0}^{x} sin (y^{2}) d y

\int_{- 2}^{x} 1 - x^{4} d x

\int_{2}^{3} \frac{x}{ln ( x - 1 )} d x

What is the integral from 1 to 2 of (1+x)^{-1} ?
The integral from 1 to 2 of (1+x)^{-1} is ln(3)-ln(2)