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

The integral of 3/(5x-1) is 3/5 ln|5x-1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 3/(5x-1)

\int \frac{3}{5 x - 1} d x

\frac{3}{5} ln ∣ 5 x - 1 ∣ + C

Solution steps

\int \frac{3}{5 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 \frac{1}{5 x - 1} d x

Apply u-substitution

= 3 \cdot \int \frac{1}{5 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}{5} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 \cdot \frac{1}{5} ln ∣ u ∣

Substitute back

u = 5 x - 1

= 3 \cdot \frac{1}{5} ln ∣ 5 x - 1 ∣

Simplify

3 \cdot \frac{1}{5} ln ∣ 5 x - 1 ∣ : \frac{3}{5} ln ∣ 5 x - 1 ∣

= \frac{3}{5} ln ∣ 5 x - 1 ∣

Add a constant to the solution

= \frac{3}{5} ln ∣ 5 x - 1 ∣ + C

\int (x^{2} - 7 x + 4) d x

\frac{d y}{d x} + \frac{y}{x} = e^{x}

\frac{d}{d x} (e^{2 x} + e^{- 3 x})

t an g e n t y = arccos (x) + x

7 x^{2} \frac{d y}{d x} = \frac{d y}{d x} + 7 x e^{- y}