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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

f (x) = 3 x^{4} - x^{3} - 5 x^{2} + 7, at x = 2

\int x^{4} (7 - \frac{x ^{5}}{10})^{3} d x

\int - x sin (y) d y

\int 3 y d y

y = x, y = \frac{7.25 x}{x ^{2} + 1}

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