What is the integral of 1/(2x+6) ?

The integral of 1/(2x+6) is 1/2 ln|2x+6|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(2x+6)

\int \frac{1}{2 x + 6} d x

\frac{1}{2} ln ∣ 2 x + 6 ∣ + C

Solution steps

\int \frac{1}{2 x + 6} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \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 \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x + 6

= \frac{1}{2} ln ∣ 2 x + 6 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 x + 6 ∣ + C

\int cos (p i x) d x

\frac{d y}{d x} + 4 y = x e^{- 4 x}

\frac{d}{d x} (4 - ∣ x ∣)

\int \frac{e ^{3 x}}{( 1 - 2 e ^{3 x} ) ^{2}} d x

\frac{d}{d x} (1 - (1 + x) e^{- x})