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

The integral from 0 to 4 of (2/(x+1)) is 2ln(5)

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 (2/(x+1))

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

2 ln (5)

Decimal

3.21887 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

: \int_{1}^{5} \frac{1}{u} d u

= 2 \cdot \int_{1}^{5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 [ln ∣ u ∣]_{1}^{5}

Compute the boundaries:

ln (5)

= 2 ln (5)

\int_{2}^{5} (x^{2}) d x

\int_{5}^{8} 10 lo g_{4} (x) d x

\int_{6}^{- 6} (3 x + 2)^{2} d x

\int_{200}^{300} - 0.1 x + 40 d x

\int_{0}^{5} 2 e^{x} + 4 cos (x) d x

What is the integral from 0 to 4 of (2/(x+1)) ?
The integral from 0 to 4 of (2/(x+1)) is 2ln(5)