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

The integral of (x^2-4)/(x^2+4) is -4arctan(x/2)+x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^2-4)/(x^2+4)

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

- 4 arctan (\frac{x}{2}) + x + C

Solution steps

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

\frac{x ^{2} - 4}{x ^{2} + 4} = - \frac{8}{x ^{2} + 4} + 1

= \int - \frac{8}{x ^{2} + 4} + 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 \frac{8}{x ^{2} + 4} d x + \int 1 d x

\int \frac{8}{x ^{2} + 4} d x = 4 arctan (\frac{x}{2})

\int 1 d x = x

= - 4 arctan (\frac{x}{2}) + x

Add a constant to the solution

= - 4 arctan (\frac{x}{2}) + x + C

What is the integral of (x^2-4)/(x^2+4) ?
The integral of (x^2-4)/(x^2+4) is -4arctan(x/2)+x+C