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 -...

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

You should notice that if you differentiate `x^2-4x+6` yields `2x - 4,` hence, you need to multiply and divide by two and then subtract and add 4, such that:

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

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

You need to use substitution to solve `(1/2)int ((2x-4))//((x^2-4x+6)^2) dx` such that:

`x^2-4x+6 = t => (2x-4)dx = dt`

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

`Put (x - 2)/(2sqrt2)=t => (dx)/(2sqrt2) = dt`

`3int 1/((x^2-4x+6)^2)dx = 3(sqrt2/4)int 1/((t^2+1)^2)`

Put `t = tan alpha => dt = 1/(cos^2 alpha) d alpha`

`3(sqrt2/4)int 1/((t^2+1)^2) = 3(sqrt2/4)int cos^2 alpha d alpha`

Use the half angle identity:

`cos^2 alpha = (1 + cos 2 alpha)/2`

`3(sqrt2/4)int cos^2 alpha d alpha = 3(sqrt2/4)int(1 + cos 2 alpha)/2 d alpha`

`3(sqrt2/8)int d alpha + 3(sqrt2/8) int cos 2 alpha d alpha`

`3(sqrt2/8)int d alpha + 3(sqrt2/16) sin 2 alpha`

`alpha = arctan((x-2)/(sqrt2))`

`3(sqrt2/8)arctan((x-2)/(sqrt2)) + 3(sqrt2/16) sin 2 (arctan((x-2)/(sqrt2))) + c`

**Hence, evaluating the integral yields `int (x^2-3x+7)/((x^2-4x+6)^2)dx =sqrt2/2 arctan ((x - 2)/sqrt2) -1/(2(x^2-4x+6)) +3(sqrt2/8)arctan((x-2)/(sqrt2)) + 3(sqrt2/16) sin 2 (arctan((x-2)/(sqrt2))) + c` **