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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(3x+3)

\int \frac{1}{3 x + 3} d x

\frac{1}{3} ln ∣ 3 x + 3 ∣ + C

Solution steps

\int \frac{1}{3 x + 3} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 3 x + 3

= \frac{1}{3} ln ∣ 3 x + 3 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 3 x + 3 ∣ + C

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

\int (x - 1) (x - 2) d x

\int ∣ 2 - x ∣ d x

\int sec (x) tan^{5} (x) d x

\int \frac{3 x + 4}{( x ^{2} + 9 )} d x