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

The integral from 2 to 5 of 1/(x+1) 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 2 to 5 of 1/(x+1)

\int_{2}^{5} \frac{1}{x + 1} d x

ln (2)

Decimal

0.69314 \dots

Solution steps

\int_{2}^{5} \frac{1}{x + 1} d x

Apply u-substitution

: \int_{3}^{6} \frac{1}{u} d u

= \int_{3}^{6} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{3}^{6}

Compute the boundaries:

ln (6) - ln (3)

= ln (6) - ln (3)

Simplify

= ln (2)

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + 2 y ^{2}})

\int \frac{81 - x ^{2}}{x ^{2}} d x

y = 3 x^{4} - 12 x^{3}

x \to - 1 lim (x)

f (x) = \frac{5 x ^{\frac{3}{2}} - 1 5 ^{\frac{1}{2}}}{2}

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