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

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

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

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

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

Expand

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

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

= 4 (\frac{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{( 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

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

\frac{d}{d x} (g (x) (x)^{2})

y = 2 sin (x), at x = \frac{3}{4} π

h (x) = \frac{3 x ^{3} - 3}{ln ( x )}

f (x) = 1 - 2 x + x^{2}

f (T) = T

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