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

The integral of (1-x-x^2)/((x^2+1)^2) is x/(x^2+1)+1/(2(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 (1-x-x^2)/((x^2+1)^2)

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int (ln (x) + cos (x)) d x

\int x + 3 x d x

\int x (x + 6)^{6} d x

\int \frac{sin ( x )}{sin ^{2} ( x )} d x

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

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