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

The integral of (x(1-x))/(1+x^2) is -x+1/2 ln|x^2+1|+arctan(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Long division

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

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

\int 1 d x = x

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

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

Add a constant to the solution

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

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