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

\int \frac{1}{4 t ^{2} + 1} d t

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

\int (sec^{2} (x) \cdot tan (x)) d x

\int \frac{1}{y ^{2} + 10 y + 30} d y

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

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