What is the integral from 2 to 5 of 3/(2x-1) ?

The integral from 2 to 5 of 3/(2x-1) is 3/2 ln(3)

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 3/(2x-1)

\int_{2}^{5} \frac{3}{2 x - 1} d x

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

Decimal

1.64791 \dots

Solution steps

\int_{2}^{5} \frac{3}{2 x - 1} d x

Take the constant out:

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

= 3 \cdot \int_{2}^{5} \frac{1}{2 x - 1} d x

Apply u-substitution

: \int_{3}^{9} \frac{1}{2 u} d u

= 3 \cdot \int_{3}^{9} \frac{1}{2 u} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int_{3}^{9} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{2} [ln ∣ u ∣]_{3}^{9}

Simplify

3 \cdot \frac{1}{2} [ln ∣ u ∣]_{3}^{9} : \frac{3}{2} [ln ∣ u ∣]_{3}^{9}

= \frac{3}{2} [ln ∣ u ∣]_{3}^{9}

Compute the boundaries:

ln (3)

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

\int_{0}^{\frac{π}{6}} \frac{cos ( x )}{x} d x

\int_{- 2}^{\infty} e^{- 0.5 x} d x

\int_{1}^{e^{5}} \frac{ln ( x )}{x} d x

\int_{- 5}^{5} (6 - ∣ x ∣) d x

\int_{0}^{\frac{π}{2}} (x - 1) cos (x) d x

What is the integral from 2 to 5 of 3/(2x-1) ?
The integral from 2 to 5 of 3/(2x-1) is 3/2 ln(3)