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

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

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

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

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 )}) + C

Solution steps

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

Take the partial fraction of

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

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

= 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 )})

Add a constant to the solution

= 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 )}) + C

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

\frac{d y}{d x} = 3 y + 4

\int cos^{4} (7 x + 2) d x

y = 6 e^{x} + x, (0, 6)

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

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