What is the integral of (2x^2-5x)/((x^2+2)^2) ?

The integral of (2x^2-5x)/((x^2+2)^2) is 1/(sqrt(2))arctan(x/(sqrt(2)))-x/(2+x^2)+5/(2(x^2+2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (2x^2-5x)/((x^2+2)^2)

\int \frac{2 x ^{2} - 5 x}{( x ^{2} + 2 ) ^{2}} d x

\frac{1}{2} arctan (\frac{x}{2}) - \frac{x}{2 + x ^{2}} + \frac{5}{2 ( x ^{2} + 2 )} + C

Solution steps

\int \frac{2 x ^{2} - 5 x}{( x ^{2} + 2 ) ^{2}} d x

Expand

\frac{2 x ^{2} - 5 x}{( x ^{2} + 2 ) ^{2}} : \frac{2 x ^{2}}{( x ^{2} + 2 ) ^{2}} - \frac{5 x}{( x ^{2} + 2 ) ^{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{2 x ^{2}}{( x ^{2} + 2 ) ^{2}} d x - \int \frac{5 x}{( x ^{2} + 2 ) ^{2}} d x

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

\int \frac{5 x}{( x ^{2} + 2 ) ^{2}} d x = - \frac{5}{2 ( x ^{2} + 2 )}

= \frac{1}{2} arctan (\frac{x}{2}) - \frac{x}{2 + x ^{2}} - (- \frac{5}{2 ( x ^{2} + 2 )})

Simplify

= \frac{1}{2} arctan (\frac{x}{2}) - \frac{x}{2 + x ^{2}} + \frac{5}{2 ( x ^{2} + 2 )}

Add a constant to the solution

= \frac{1}{2} arctan (\frac{x}{2}) - \frac{x}{2 + x ^{2}} + \frac{5}{2 ( x ^{2} + 2 )} + C

What is the integral of (2x^2-5x)/((x^2+2)^2) ?
The integral of (2x^2-5x)/((x^2+2)^2) is 1/(sqrt(2))arctan(x/(sqrt(2)))-x/(2+x^2)+5/(2(x^2+2))+C