What is the integral of 1/(-4x+2) ?

The integral of 1/(-4x+2) is -1/4 ln|-4x+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/(-4x+2)

\int \frac{1}{- 4 x + 2} d x

- \frac{1}{4} ln ∣ - 4 x + 2 ∣ + C

Solution steps

\int \frac{1}{- 4 x + 2} d x

Apply u-substitution

: \int - \frac{1}{4 u} d u

= \int - \frac{1}{4 u} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{4} ln ∣ u ∣

Substitute back

u = - 4 x + 2

= - \frac{1}{4} ln ∣ - 4 x + 2 ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ - 4 x + 2 ∣ + C

\int \frac{1}{9 Z ^{2} + 4} d Z

\int e^{x} sin (π) x d x

\int \frac{r}{1} d r

\int \frac{ln ( 1 + x )}{x} d x

\int x \cdot arctan^{2} (x) d x

What is the integral of 1/(-4x+2) ?
The integral of 1/(-4x+2) is -1/4 ln|-4x+2|+C