What is the integral from 0 to 4 of 1/(x+2) ?

The integral from 0 to 4 of 1/(x+2) is ln(3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 4 of 1/(x+2)

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

ln (3)

Decimal Notation

1.09861 \dots

Solution steps

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

Apply u-substitution

: \int_{2}^{6} \frac{1}{u} d u

= \int_{2}^{6} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{2}^{6}

Compute the boundaries:

ln (6) - ln (2)

= ln (6) - ln (2)

Simplify

= ln (3)

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + 2 y = 0

\int \frac{2 z}{1 - z ^{2}} d z

x \to 0 + lim (\frac{4}{x ^{2}})

\int 2 x y^{3} z^{4} d x

\int \frac{y - 2}{y + 3} d y

What is the integral from 0 to 4 of 1/(x+2) ?
The integral from 0 to 4 of 1/(x+2) is ln(3)