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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

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

\int 90 x d x

\int x arccos (\frac{1}{x}) d x

\int \frac{1}{x ( 9 - x ^{2} )} d x

\int i n^{3} x d x

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