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

The integral from 0 to 3 of (-2)/(5x+1) is -8/5 ln(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- \frac{8}{5} ln (2)

Decimal Notation

- 1.10903 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{1}^{16} \frac{1}{5 u} d u

= - 2 \cdot \int_{1}^{16} \frac{1}{5 u} d u

Take the constant out:

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

= - 2 \cdot \frac{1}{5} \cdot \int_{1}^{16} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 2 \cdot \frac{1}{5} [ln ∣ u ∣]_{1}^{16}

Simplify

- 2 \cdot \frac{1}{5} [ln ∣ u ∣]_{1}^{16} : - \frac{2}{5} [ln ∣ u ∣]_{1}^{16}

= - \frac{2}{5} [ln ∣ u ∣]_{1}^{16}

Compute the boundaries:

4 ln (2)

= - \frac{2}{5} \cdot 4 ln (2)

Simplify

= - \frac{8}{5} ln (2)

\int_{- 1}^{2} 4 t^{2} - 8 t + 5 d t

\int_{1}^{0} e^{x} d x

\int_{0}^{2 π} \frac{1}{5 - 4 cos ( x )} d x

\int_{0}^{t} cos (2 wt) d t

\int_{3}^{9} 2 x - 2 - x + 5 d x

What is the integral from 0 to 3 of (-2)/(5x+1) ?
The integral from 0 to 3 of (-2)/(5x+1) is -8/5 ln(2)