What is the integral of 1/(8x-2) ?

The integral of 1/(8x-2) is 1/8 ln|8x-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/(8x-2)

\int \frac{1}{8 x - 2} d x

\frac{1}{8} ln ∣ 8 x - 2 ∣ + C

Solution steps

\int \frac{1}{8 x - 2} d x

Apply u-substitution

: \int \frac{1}{8 u} d u

= \int \frac{1}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{8} ln ∣ u ∣

Substitute back

u = 8 x - 2

= \frac{1}{8} ln ∣ 8 x - 2 ∣

Add a constant to the solution

= \frac{1}{8} ln ∣ 8 x - 2 ∣ + C

\int e^{\frac{1}{5} x} d x

\int (6 x + 1) (x^{2} + x)^{2} d x

\int u (3 u^{2} + 1) d u

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

\int x^{6} sin (x^{7} + 2) d x