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

The integral of (2-x^4)/(1+x^2) is -(x^3)/3+x+arctan(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (2-x^4)/(1+x^2)

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

- \frac{x ^{3}}{3} + x + arctan (x) + C

Solution steps

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

Expand

\frac{2 - x ^{4}}{1 + x ^{2}} : \frac{2}{1 + x ^{2}} - \frac{x ^{4}}{1 + x ^{2}}

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

\int \frac{2}{1 + x ^{2}} d x = 2 arctan (x)

\int \frac{x ^{4}}{1 + x ^{2}} d x = \frac{x ^{3}}{3} - x + arctan (x)

= 2 arctan (x) - (\frac{x ^{3}}{3} - x + arctan (x))

Simplify

= - \frac{x ^{3}}{3} + x + arctan (x)

Add a constant to the solution

= - \frac{x ^{3}}{3} + x + arctan (x) + C

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