Answer
$\frac{7\sqrt 2}{8}\tan^{-1}\left(\frac{x-2}{\sqrt 2}\right)+\frac{3x-8}{4(x^2-4x+6)}+C$
Work Step by Step
$\int\frac{x^2-3x+7}{(x^2-4x+6)^2}dx$ = $\int\left[\frac{Ax+B}{x^2-4x+6}+\frac{Cx+D}{(x^2-4x+6)^2}\right]dx$
$x^2-3x+7$ = $(Ax+B)(x^2-4x+6)+(Cx+D)$
$x^2-3x+7$ = $(Ax^3+Bx^2-4Ax^2-4Bx+6Ax+6B)+(Cx+D)$
$0$ = $A$
$1$ = $-4A+B$
$-3$ = $6A-4B+C$
$7$ = $6B+D$
$A$ = $0$, $B$ = $1$, $C$ = $1$, $D$ = $1$
so
$\int\frac{x^2-3x+7}{(x^2-4x+6)^2}dx$ = $\int\left[\frac{1}{x^2-4x+6}+\frac{x+1}{(x^2-4x+6)^2}\right]dx$ = $\int\left[\frac{1}{(x-2)^2+2}+\frac{x-2}{(x^2-4x+6)^2}+\frac{3}{(x^2-4x+6)^2}\right]dx$ = $\frac{1}{\sqrt 2}\tan^{-1}\left(\frac{x-2}{\sqrt 2}\right)-\frac{1}{2(x^2-4x+6)}+\frac{3\sqrt 2}{8}\tan^{-1}\left(\frac{x-2}{\sqrt 2}\right)+\frac{3(x-2)}{4(x^2-4x+6)}+C$ = $\frac{7\sqrt 2}{8}\tan^{-1}\left(\frac{x-2}{\sqrt 2}\right)+\frac{3x-8}{4(x^2-4x+6)}+C$
