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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

x - \frac{3}{2} arctan (x) + \frac{x}{2 ( 1 + x ^{2} )} + C

Solution steps

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

Take the partial fraction of

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

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

\int 1 d x = x

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

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

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

Simplify

x - 2 arctan (x) + \frac{1}{2} (arctan (x) + \frac{x}{1 + x ^{2}}) : x - \frac{3}{2} arctan (x) + \frac{x}{2 ( 1 + x ^{2} )}

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

Add a constant to the solution

= x - \frac{3}{2} arctan (x) + \frac{x}{2 ( 1 + x ^{2} )} + C

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

\int \frac{y}{8 - y} d y

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

\int 2 x 3 x^{2} + 5 d x

\int sec (2 x + 3) tan (2 x + 3) d x

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