What is the integral of (x^2-3x+7)/((x^2-4x+6)^2) ?

The integral of (x^2-3x+7)/((x^2-4x+6)^2) is

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integral of (x^2-3x+7)/((x^2-4x+6)^2)

\int \frac{x ^{2} - 3 x + 7}{( x ^{2} - 4 x + 6 ) ^{2}} d x

\frac{1}{2} arctan (\frac{x - 2}{2}) - \frac{1}{2 ( x ^{2} - 4 x + 6 )} + \frac{3}{4 2} (arctan (\frac{1}{2} (x - 2)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (x - 2)))) + C

Solution steps

\int \frac{x ^{2} - 3 x + 7}{( x ^{2} - 4 x + 6 ) ^{2}} d x

Take the partial fraction of

\frac{x ^{2} - 3 x + 7}{( x ^{2} - 4 x + 6 ) ^{2}} : \frac{1}{x ^{2} - 4 x + 6} + \frac{x + 1}{( x ^{2} - 4 x + 6 ) ^{2}}

= \int \frac{1}{x ^{2} - 4 x + 6} + \frac{x + 1}{( x ^{2} - 4 x + 6 ) ^{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{1}{x ^{2} - 4 x + 6} d x + \int \frac{x + 1}{( x ^{2} - 4 x + 6 ) ^{2}} d x

\int \frac{1}{x ^{2} - 4 x + 6} d x = \frac{1}{2} arctan (\frac{x - 2}{2})

\int \frac{x + 1}{( x ^{2} - 4 x + 6 ) ^{2}} d x = - \frac{1}{2 ( x ^{2} - 4 x + 6 )} + \frac{3}{4 2} (arctan (\frac{1}{2} (x - 2)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (x - 2))))

= \frac{1}{2} arctan (\frac{x - 2}{2}) - \frac{1}{2 ( x ^{2} - 4 x + 6 )} + \frac{3}{4 2} (arctan (\frac{1}{2} (x - 2)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (x - 2))))

Add a constant to the solution

= \frac{1}{2} arctan (\frac{x - 2}{2}) - \frac{1}{2 ( x ^{2} - 4 x + 6 )} + \frac{3}{4 2} (arctan (\frac{1}{2} (x - 2)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (x - 2)))) + C

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What is the integral of (x^2-3x+7)/((x^2-4x+6)^2) ?
The integral of (x^2-3x+7)/((x^2-4x+6)^2) is