What is the integral from 0 to 12 of 1/(x+12) ?

The integral from 0 to 12 of 1/(x+12) is ln(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 12 of 1/(x+12)

\int_{0}^{12} \frac{1}{x + 12} d x

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

\int_{0}^{12} \frac{1}{x + 12} d x

Apply u-substitution

: \int_{12}^{24} \frac{1}{u} d u

= \int_{12}^{24} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{12}^{24}

Compute the boundaries:

ln (24) - ln (12)

= ln (24) - ln (12)

Simplify

= ln (2)

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

\int_{0}^{5} 2 t d t

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

\int_{0}^{25} y + 11 d y

\int_{0}^{\infty} 18 e^{- 2 x} d x

What is the integral from 0 to 12 of 1/(x+12) ?
The integral from 0 to 12 of 1/(x+12) is ln(2)