`int (x^2 - 3x + 7)/(x^2 - 4x + 6)^2 dx` Evaluate the integral

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You may rebuild the structure you find at denominator, such that:

`int (x^2-3x+7)/((x^2-4x+6)^2)dx = int ((x^2-4x+6) + x + 1)/((x^2-4x+6)^2)dx`

Separate into two integrals:

`int (x^2-4x+6)/((x^2-4x+6)^2)dx + int (x + 1)/((x^2-4x+6)^2)dx`

Take the first integral and reduce like terms:

`int (x^2-4x+6)/((x^2-4x+6)^2)dx = int 1/((x^2-4x+6))dx `

You may write `x^2 - 4x + 6 = x^2 - 4x + 4 + 2 = (x- 2)^2 + 2`

`int 1/((x^2-4x+6))dx= int 1/((x- 2)^2 + 2)dx = sqrt2/2 arctan ((x - 2)/sqrt2) + c`

You need to take the integral` int (x + 1)/((x^2-4x+6)^2)dx ` and to separate it into two simpler integrals:

`int (x + 1)/((x^2-4x+6)^2)dx = int x/((x^2-4x+6)^2)dx + int...

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