What is the integral of 1/(9x+2) ?

The integral of 1/(9x+2) is 1/9 ln|9x+2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(9x+2)

\int \frac{1}{9 x + 2} d x

\frac{1}{9} ln ∣ 9 x + 2 ∣ + C

Solution steps

\int \frac{1}{9 x + 2} d x

Apply u-substitution

: \int \frac{1}{9 u} d u

= \int \frac{1}{9 u} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{9} ln ∣ u ∣

Substitute back

u = 9 x + 2

= \frac{1}{9} ln ∣ 9 x + 2 ∣

Add a constant to the solution

= \frac{1}{9} ln ∣ 9 x + 2 ∣ + C

\int - 4 x cos (\frac{x}{2}) d x

\int 4 x - 6 x^{2} d x

\int x \cdot tan (\frac{x}{3}) d x

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

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