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

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

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

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

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)) + 2 ln x^{2} - 2 x + 2 + 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{x ^{4} - 2 x ^{2} + 2}{( x ^{2} - 2 x + 2 ) ^{2}} d x

Take the partial fraction of

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

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

\int 1 d x = x

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

= x + 2 ln x^{2} - 2 x + 2 + 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

x + 2 ln x^{2} - 2 x + 2 + 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)) : 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)) + 2 ln x^{2} - 2 x + 2 + 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}

= 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)) + 2 ln x^{2} - 2 x + 2 + 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

= 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)) + 2 ln x^{2} - 2 x + 2 + 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

\int \frac{e ^{1 + l n (x)}}{x} d x

y^{^{'}} = \frac{y ^{2} + 1}{x}

x \to \infty lim (I n (x))

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

\frac{\partial}{\partial x} (3 x^{3} + 4 x^{2} y - y^{3})

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