What is the integral of (x^2+1)/((x^2-2x+2)^2) ?

The integral of (x^2+1)/((x^2-2x+2)^2) is

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integral of (x^2+1)/((x^2-2x+2)^2)

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

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

Solution steps

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

Expand

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{x - 13}{x - 13}

\int_{- 5}^{5} \frac{1}{2} (25 - x^{2}) d x

(6 x + 1) y^{2} \frac{d y}{d x} + 4 x^{2} + 2 y^{3} = 0

y = \frac{x ^{2}}{x + 5}, (1, \frac{1}{6})

\frac{\partial}{\partial x} (\frac{1}{x} + \frac{1}{y + z})

What is the integral of (x^2+1)/((x^2-2x+2)^2) ?
The integral of (x^2+1)/((x^2-2x+2)^2) is