What is the integral of (3x^2+6)/((x^2-2x+2)^2) ?

The integral of (3x^2+6)/((x^2-2x+2)^2) is

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integral of (3x^2+6)/((x^2-2x+2)^2)

\int \frac{3 x ^{2} + 6}{( x ^{2} - 2 x + 2 ) ^{2}} d x

3 arctan (x - 1) + 6 (\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{3 x ^{2} + 6}{( x ^{2} - 2 x + 2 ) ^{2}} d x

Take the partial fraction of

\frac{3 x ^{2} + 6}{( x ^{2} - 2 x + 2 ) ^{2}} : \frac{3}{x ^{2} - 2 x + 2} + \frac{6 x}{( x ^{2} - 2 x + 2 ) ^{2}}

= \int \frac{3}{x ^{2} - 2 x + 2} + \frac{6 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{3}{x ^{2} - 2 x + 2} d x + \int \frac{6 x}{( x ^{2} - 2 x + 2 ) ^{2}} d x

\int \frac{3}{x ^{2} - 2 x + 2} d x = 3 arctan (x - 1)

\int \frac{6 x}{( x ^{2} - 2 x + 2 ) ^{2}} d x = 6 (\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 )})

= 3 arctan (x - 1) + 6 (\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

= 3 arctan (x - 1) + 6 (\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{d}{d x} (sin (x^{3}) + 3 x y)

f (x) = \frac{x}{x + 3}, at x = 1

f (x) = e^{6 x} cos (9 x)

\frac{d}{d x} (ln (2 + x^{2}))

\int cos (5 x) d x

What is the integral of (3x^2+6)/((x^2-2x+2)^2) ?
The integral of (3x^2+6)/((x^2-2x+2)^2) is