What is the integral from 2 to 3 of 1/(2x-3) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 2 to 3 of 1/(2x-3)

\int_{2}^{3} \frac{1}{2 x - 3} d x

\frac{1}{2} ln (3)

Decimal

0.54930 \dots

Solution steps

\int_{2}^{3} \frac{1}{2 x - 3} d x

Apply u-substitution

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} [ln ∣ u ∣]_{1}^{3}

Compute the boundaries:

ln (3)

= \frac{1}{2} ln (3)

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