What is the integral of ((x-3)^3)/(x^2-9) ?

The integral of ((x-3)^3)/(x^2-9) is (x^2)/2-9x+36ln|x+3|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of ((x-3)^3)/(x^2-9)

\int \frac{( x - 3 ) ^{3}}{x ^{2} - 9} d x

\frac{x ^{2}}{2} - 9 x + 36 ln ∣ x + 3 ∣ + C

Solution steps

\int \frac{( x - 3 ) ^{3}}{x ^{2} - 9} d x

Long division

\frac{( x - 3 ) ^{3}}{x ^{2} - 9} : x - 9 + \frac{36}{x + 3}

= \int x - 9 + \frac{36}{x + 3} 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 x d x - \int 9 d x + \int \frac{36}{x + 3} d x

\int x d x = \frac{x ^{2}}{2}

\int 9 d x = 9 x

\int \frac{36}{x + 3} d x = 36 ln ∣ x + 3 ∣

= \frac{x ^{2}}{2} - 9 x + 36 ln ∣ x + 3 ∣

Add a constant to the solution

= \frac{x ^{2}}{2} - 9 x + 36 ln ∣ x + 3 ∣ + C

What is the integral of ((x-3)^3)/(x^2-9) ?
The integral of ((x-3)^3)/(x^2-9) is (x^2)/2-9x+36ln|x+3|+C