If the integrand is
[tex](x^2-3x+12)(x^2-4x+11)^2[/tex]
then the easiest way to compute this is to expand the product completely, then integrate the resulting polynomial.
If instead it's
[tex]\dfrac{x^2-3x+12}{(x^2-4x+11)^2}[/tex]
you can split the integrand into partial fractions
[tex]\dfrac{x^2-3x+12}{(x^2-4x+11)^2}=\dfrac{a_0+a_1x}{x^2-4x+11}+\dfrac{b_0+b_1x}{(x^2-4x+11)^2}[/tex]
[tex]\implies x^2-3x+12=(a_0+a_1x)(x^2-4x+11)+b_0+b_1x[/tex]
On the right side, we have
[tex]a_1x^3+(a_0-4a_1)x^2+(-4a_0+11a_1+b_1)x+(11a_0+b_0)[/tex]
so that
[tex]\begin{cases}a_1=0\\a_0-4a_1=1\implies a_0=1\\-4a_0+11a_1+b_1=-3\implies b_1=1\\11a_0+b_0=12\implies b_0=1\end{cases}[/tex]
Then
[tex]\displaystyle\int\frac{x^2-3x+12}{(x^2-4x+11)^2}\,\mathrm dx=\int\left(\frac x{x^2-4x+11}+\frac{x+1}{(x^2-4x+11)^2}\right)\,\mathrm dx[/tex]
In both denominators, we can complete the square to write
[tex]x^2-4x+11=(x-2)^2+7[/tex]
[tex]\displaystyle\int\left(\frac x{(x-2)^2+7}+\frac{x+1}{((x-2)^2+7)^2}\right)\,\mathrm dx[/tex]
Then substitute
[tex]x-2=\sqrt7\tan t\implies\mathrm dx=\sqrt7\sec^2t\,\mathrm dt[/tex]
[tex]\implies\displaystyle\sqrt7\int\left(\frac{\sqrt7\tan t+2}{7\tan^2t+7}+\frac{\sqrt7\tan t+3}{(7\tan^2t+7)^2}\right)\sec^2t\,\mathrm dt[/tex]
We have
[tex]\tan^2t+1=\sec^2t[/tex]
[tex]\implies\displaystyle\frac1{\sqrt7}\int(\sqrt7\tan t+2)\,\mathrm dt+\frac1{7\sqrt7}\int\frac{\sqrt7\tan t+3}{\sec^2t}\,\mathrm dt[/tex]
[tex]\implies\displaystyle\int\tan t\,\mathrm dt+\frac2{\sqrt7}\int\mathrm dt+\frac17\int\frac{\tan t}{\sec^2t}\,\mathrm dt+\frac3{7\sqrt7}\int\cos^2t\,\mathrm dt[/tex]
Since
[tex]\dfrac{\tan t}{\sec^2t}=\dfrac{\frac{\sin t}{\cos t}}{\frac1{\cos^2t}}=\sin t\cos t=\dfrac12\sin2t[/tex]
and
[tex]\cos^2t=\dfrac{1+\cos2t}2[/tex]
all the remaining integrals are trivial; we end up with
[tex]-\ln|\cos t|+\dfrac{31}{14\sqrt7}t-\dfrac1{28}\cos2t+\dfrac3{28\sqrt7}\sin2t+C[/tex]
Reverse the substitution:
[tex]t=\tan^{-1}\dfrac{x-2}{\sqrt7}\implies\begin{cases}\cos t=\sqrt{\frac7{x^2-4x+11}}\\\cos2t=-\frac{x^2-4x-3}{x^2-4x+11}\\\sin2t=\frac{2\sqrt7(x-2)}{x^2-4x+11}\end{cases}[/tex]
[tex]\implies\dfrac12\ln(x^2-4x+11)+\dfrac{31}{14\sqrt7}\tan^{-1}\dfrac{x-2}{\sqrt7}+\dfrac1{28}\dfrac{x^2+2x-15}{x^2-4x+11}+C[/tex]
