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

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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

(8.7) (0.3)

\int \frac{2}{25 - 100 x ^{2}} d x

y = \frac{arctan ( 2 x )}{x}

\frac{\partial}{\partial x} (\frac{x ^{2} + y ^{2}}{z ^{2}})

\int_{2}^{3} \frac{2 x}{( x ^{2} + 1 ) ^{3}} d x

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