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

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

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

\frac{1}{2} ln (5)

Decimal

0.80471 \dots

Solution steps

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

Apply u-substitution

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{5}

Compute the boundaries:

ln (5)

= \frac{1}{2} ln (5)

\int_{- 1}^{3} x^{2} (x - 2) d x

\int_{0}^{π} (1 + x^{2}) cos (x) d x

\int_{- 2}^{2} 7 x + 5 d x

\int_{4}^{5} x e^{- x} d x

\int_{- 1}^{4} ∣ x - 2 ∣ d x

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