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

The integral of (x^3+4x)/(x^2-1) is 1/2 x^2-1/2+5/2 ln|x^2-1|+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+4x)/(x^2-1)

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

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

Solution steps

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

Expand

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

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{31 x - 33}{2 x ^{2} - 6 x} d x

\int 6 cosh (\frac{x}{2} - ln (3)) d x

\int 7 x sin (8 x) d x

\int 14 x (7 x^{2} + 7)^{7} d x

\int 12 x - 2 d x

What is the integral of (x^3+4x)/(x^2-1) ?
The integral of (x^3+4x)/(x^2-1) is 1/2 x^2-1/2+5/2 ln|x^2-1|+C