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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Add a constant to the solution

= 4 arctan (x) - \frac{1}{2 ( x ^{2} + 1 )} + C

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