What is the integral from 0 to 1 of 3/(2x-5) ?

The integral from 0 to 1 of 3/(2x-5) is 3/2 (ln(3)-ln(5))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 1 of 3/(2x-5)

\int_{0}^{1} \frac{3}{2 x - 5} d x

\frac{3}{2} (ln (3) - ln (5))

Decimal

- 0.76623 \dots

Solution steps

\int_{0}^{1} \frac{3}{2 x - 5} d x

Take the constant out:

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

= 3 \cdot \int_{0}^{1} \frac{1}{2 x - 5} d x

Apply u-substitution

= 3 \cdot \int_{- 5}^{- 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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{2} [ln ∣ u ∣]_{- 5}^{- 3}

Simplify

3 \cdot \frac{1}{2} [ln ∣ u ∣]_{- 5}^{- 3} : \frac{3}{2} [ln ∣ u ∣]_{- 5}^{- 3}

= \frac{3}{2} [ln ∣ u ∣]_{- 5}^{- 3}

Compute the boundaries:

ln (3) - ln (5)

= \frac{3}{2} (ln (3) - ln (5))

What is the integral from 0 to 1 of 3/(2x-5) ?
The integral from 0 to 1 of 3/(2x-5) is 3/2 (ln(3)-ln(5))