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

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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

g (x) = \frac{6 x ^{2}}{2 - x}

\int sin^{4} (3 x) - cos^{4} (3 x) d x

\int e^{- s t} \cdot sin (t) d t

\int \frac{1}{x ^{\frac{1}{3}} + 1} d x

\frac{d y}{d x} = x e^{2 y}

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