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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(x(x-4))

\int \frac{1}{x ( x - 4 )} d x

- \frac{1}{4} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{1}{x ( x - 4 )} d x

Take the partial fraction of

\frac{1}{x ( x - 4 )} : - \frac{1}{4 x} + \frac{1}{4 ( x - 4 )}

= \int - \frac{1}{4 x} + \frac{1}{4 ( x - 4 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{4 x} d x + \int \frac{1}{4 ( x - 4 )} d x

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

\int \frac{1}{4 ( x - 4 )} d x = \frac{1}{4} ln ∣ x - 4 ∣

= - \frac{1}{4} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 4 ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ x ∣ + \frac{1}{4} ln ∣ x - 4 ∣ + C

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