What is the integral of 1/(9x-4) ?

The integral of 1/(9x-4) is 1/9 ln|9x-4|+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-4)

\int \frac{1}{9 x - 4} d x

\frac{1}{9} ln ∣ 9 x - 4 ∣ + C

Solution steps

\int \frac{1}{9 x - 4} 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 - 4

= \frac{1}{9} ln ∣ 9 x - 4 ∣

Add a constant to the solution

= \frac{1}{9} ln ∣ 9 x - 4 ∣ + C

\int \frac{x ^{2} - 2 x}{x ^{3} - 3 x ^{2} + 1} d x

\frac{d}{d x} (e^{s i n (x)} \cdot cos (sin (x)))

y^{^{'}} - \frac{8}{x} y = \frac{y ^{4}}{x ^{18}}

\int X^{2} d X

\frac{\partial}{\partial x} (2 y^{3} + 4 x)