What is the integral of (2x^2+6)/((x^2-2x+2)^2) ?

The integral of (2x^2+6)/((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+6)/((x^2-2x+2)^2)

\int \frac{2 x ^{2} + 6}{( 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)) + 3 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} + C

Solution steps

\int \frac{2 x ^{2} + 6}{( x ^{2} - 2 x + 2 ) ^{2}} d x

Take the partial fraction of

\frac{2 x ^{2} + 6}{( x ^{2} - 2 x + 2 ) ^{2}} : \frac{2}{x ^{2} - 2 x + 2} + \frac{4 x + 2}{( x ^{2} - 2 x + 2 ) ^{2}}

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

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

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

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

Simplify

2 arctan (x - 1) + 4 (\frac{x ( arctan ( x - 1 ) ( x ^{2} - 2 x + 2 ) + x - 1 )}{2 ( x ^{2} - 2 x + 2 )} - \frac{( x - 1 ) ( arctan ( x - 1 ) + \frac{x - 1}{x ^{2} - 2 x + 2} ) ( x ^{2} - 2 x + 2 ) + 1}{2 ( 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)) + 3 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}

= \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)) + 3 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}

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)) + 3 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} + C

x \to \infty lim (ln (2 x))

x = - 1, x = 3, y = x^{3} - 1, y = 0

\frac{d}{d x} (x^{8} ln (5 x))

\frac{d}{d x} (1200 x^{2} - 0.2 x)

x \to 2 lim (\frac{x ^{2}}{x + 2})

What is the integral of (2x^2+6)/((x^2-2x+2)^2) ?
The integral of (2x^2+6)/((x^2-2x+2)^2) is