What is the integral of 1/((x-6)(x-5)) ?

The integral of 1/((x-6)(x-5)) is ln|x-6|-ln|x-5|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/((x-6)(x-5))

\int \frac{1}{( x - 6 ) ( x - 5 )} d x

ln ∣ x - 6 ∣ - ln ∣ x - 5 ∣ + C

Solution steps

\int \frac{1}{( x - 6 ) ( x - 5 )} d x

Take the partial fraction of

\frac{1}{( x - 6 ) ( x - 5 )} : \frac{1}{x - 6} - \frac{1}{x - 5}

= \int \frac{1}{x - 6} - \frac{1}{x - 5} 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}{x - 6} d x - \int \frac{1}{x - 5} d x

\int \frac{1}{x - 6} d x = ln ∣ x - 6 ∣

\int \frac{1}{x - 5} d x = ln ∣ x - 5 ∣

= ln ∣ x - 6 ∣ - ln ∣ x - 5 ∣

Add a constant to the solution

= ln ∣ x - 6 ∣ - ln ∣ x - 5 ∣ + C

What is the integral of 1/((x-6)(x-5)) ?
The integral of 1/((x-6)(x-5)) is ln|x-6|-ln|x-5|+C