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

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

integral

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

\frac{x ^{3}}{3} + \frac{1}{4} ln ∣ x + 2 ∣ + \frac{3}{4} ln ∣ x - 2 ∣ + C

Solution steps

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

Long division

\frac{x ^{4} - 4 x ^{2} + x + 1}{x ^{2} - 4} : x^{2} + \frac{x + 1}{x ^{2} - 4}

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

\int x^{2} d x = \frac{x ^{3}}{3}

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

= \frac{x ^{3}}{3} + \frac{1}{4} ln ∣ x + 2 ∣ + \frac{3}{4} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{1}{4} ln ∣ x + 2 ∣ + \frac{3}{4} ln ∣ x - 2 ∣ + C

\frac{2 y}{x}

e^{- 2} (x)^{2}

\frac{2 x}{3}

\frac{( t + 1 )}{t}

(e^{- 2} x) sinh^{2} (x)

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