What is the integral of (x^2-5x-4)/((x^2-9)(x+2)) ?

The integral of (x^2-5x-4)/((x^2-9)(x+2)) is 10/3 ln|x+3|-1/3 ln|x-3|-2ln|x+2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2-5x-4)/((x^2-9)(x+2))

\int \frac{x ^{2} - 5 x - 4}{( x ^{2} - 9 ) ( x + 2 )} d x

\frac{10}{3} ln ∣ x + 3 ∣ - \frac{1}{3} ln ∣ x - 3 ∣ - 2 ln ∣ x + 2 ∣ + C

Solution steps

\int \frac{x ^{2} - 5 x - 4}{( x ^{2} - 9 ) ( x + 2 )} d x

Take the partial fraction of

\frac{x ^{2} - 5 x - 4}{( x ^{2} - 9 ) ( x + 2 )} : \frac{10}{3 ( x + 3 )} - \frac{1}{3 ( x - 3 )} - \frac{2}{x + 2}

= \int \frac{10}{3 ( x + 3 )} - \frac{1}{3 ( x - 3 )} - \frac{2}{x + 2} 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{10}{3 ( x + 3 )} d x - \int \frac{1}{3 ( x - 3 )} d x - \int \frac{2}{x + 2} d x

\int \frac{10}{3 ( x + 3 )} d x = \frac{10}{3} ln ∣ x + 3 ∣

\int \frac{1}{3 ( x - 3 )} d x = \frac{1}{3} ln ∣ x - 3 ∣

\int \frac{2}{x + 2} d x = 2 ln ∣ x + 2 ∣

= \frac{10}{3} ln ∣ x + 3 ∣ - \frac{1}{3} ln ∣ x - 3 ∣ - 2 ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{10}{3} ln ∣ x + 3 ∣ - \frac{1}{3} ln ∣ x - 3 ∣ - 2 ln ∣ x + 2 ∣ + C

\frac{d}{d x} (4 - \frac{9}{2 x ^{\frac{1}{10}}})

\frac{d}{d x} (\frac{5}{x - 4})

y^{^{''}} + y = csc (x)

\int \frac{x ^{4} - x ^{3} + x + 1}{x ^{3} + x ^{2}} d x

\int \frac{x ^{3} + 3 x ^{2} + 3 x + 63}{( x ^{2} - 9 ) ^{2}} d x

What is the integral of (x^2-5x-4)/((x^2-9)(x+2)) ?
The integral of (x^2-5x-4)/((x^2-9)(x+2)) is 10/3 ln|x+3|-1/3 ln|x-3|-2ln|x+2|+C