What is the integral of (x^3+2x+5)/(x^2-1) ?

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

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

\frac{1}{2} x^{2} - \frac{1}{2} + \frac{3}{2} ln x^{2} - 1 - \frac{5}{2} ln ∣ x + 1 ∣ + \frac{5}{2} ln ∣ x - 1 ∣ + C

Solution steps

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

Expand

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

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

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

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

\int \frac{5}{x ^{2} - 1} d x = - 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

= \frac{1}{2} (x^{2} - 1 + ln x^{2} - 1) + ln x^{2} - 1 - 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

Simplify

\frac{1}{2} (x^{2} - 1 + ln x^{2} - 1) + ln x^{2} - 1 - 5 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) : \frac{1}{2} x^{2} - \frac{1}{2} + \frac{3}{2} ln x^{2} - 1 - \frac{5}{2} ln ∣ x + 1 ∣ + \frac{5}{2} ln ∣ x - 1 ∣

= \frac{1}{2} x^{2} - \frac{1}{2} + \frac{3}{2} ln x^{2} - 1 - \frac{5}{2} ln ∣ x + 1 ∣ + \frac{5}{2} ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{1}{2} x^{2} - \frac{1}{2} + \frac{3}{2} ln x^{2} - 1 - \frac{5}{2} ln ∣ x + 1 ∣ + \frac{5}{2} ln ∣ x - 1 ∣ + C

\int arccoth (x) d x

\int x^{2} cos (4 x^{3} - 1) d x

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

\int \frac{6 x ^{2} + x - 3}{x ^{3} - x} d x

\int \frac{1}{9 t ^{2} + 4} d t

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