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

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

\int_{0}^{3} \frac{1}{x + 1} d x

2 ln (2)

Decimal Notation

1.38629 \dots

Solution steps

\int_{0}^{3} \frac{1}{x + 1} d x

Apply u-substitution

: \int_{1}^{4} \frac{1}{u} d u

= \int_{1}^{4} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{4}

Compute the boundaries:

2 ln (2)

= 2 ln (2)

y = \frac{7}{x + 5}

\int (x - 4)^{2} d x

x \to 0 + lim ((x + 1)^{I n x})

\int \frac{( ln ( x ) )}{x ^{3}} d x

\frac{d v}{d t} = - k v

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