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

The integral of (x^4)/((x^2-1)(x+2)) is (x^2)/2-2x-1/2 ln|x+1|+1/6 ln|x-1|+16/3 ln|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^4)/((x^2-1)(x+2))

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

\frac{x ^{2}}{2} - 2 x - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x - 1 ∣ + \frac{16}{3} ln ∣ x + 2 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{x ^{4}}{( x ^{2} - 1 ) ( x + 2 )} : x - 2 - \frac{1}{2 ( x + 1 )} + \frac{1}{6 ( x - 1 )} + \frac{16}{3 ( x + 2 )}

= \int x - 2 - \frac{1}{2 ( x + 1 )} + \frac{1}{6 ( x - 1 )} + \frac{16}{3 ( 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 x d x - \int 2 d x - \int \frac{1}{2 ( x + 1 )} d x + \int \frac{1}{6 ( x - 1 )} d x + \int \frac{16}{3 ( x + 2 )} d x

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

\int 2 d x = 2 x

\int \frac{1}{2 ( x + 1 )} d x = \frac{1}{2} ln ∣ x + 1 ∣

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

\int \frac{16}{3 ( x + 2 )} d x = \frac{16}{3} ln ∣ x + 2 ∣

= \frac{x ^{2}}{2} - 2 x - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x - 1 ∣ + \frac{16}{3} ln ∣ x + 2 ∣

Add a constant to the solution

= \frac{x ^{2}}{2} - 2 x - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{6} ln ∣ x - 1 ∣ + \frac{16}{3} ln ∣ x + 2 ∣ + C

\int (4 x + 7)^{11} d x

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

\int \frac{2 x ^{2} + x}{( 4 x ^{3} + 3 x ^{2} ) ^{2}} d x

\int e^{x} - 4 cos (x) d x

\int (\frac{( 1 - x )}{x})^{2} d x

What is the integral of (x^4)/((x^2-1)(x+2)) ?
The integral of (x^4)/((x^2-1)(x+2)) is (x^2)/2-2x-1/2 ln|x+1|+1/6 ln|x-1|+16/3 ln|x+2|+C