What is the integral of (x^5+2)/(x^2-1) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^5+2)/(x^2-1)

\int \frac{x ^{5} + 2}{x ^{2} - 1} d x

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

Solution steps

\int \frac{x ^{5} + 2}{x ^{2} - 1} d x

Long division

\frac{x ^{5} + 2}{x ^{2} - 1} : x^{3} + x + \frac{x + 2}{x ^{2} - 1}

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

\int x^{3} d x = \frac{x ^{4}}{4}

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (sin (x) \cdot cos (x))

y^{^{''}} + 9 y = 2 cos (3 x), y (0) = 1, y^{^{'}} (0) = 0

\frac{\partial}{\partial x} (x^{2} \cdot e^{- 2 y})

\frac{\partial}{\partial x} (3 y^{2} x)

\frac{\partial}{\partial x} (6 x^{7 y})

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