If the integral is simply
[tex]\displaystyle\int(x^3-6x^2+9x+3)(3x^2-12x+9)\,\mathrm dx[/tex]
then notice that
[tex]\mathrm d(x^3-6x^2+9x+3)=(3x^2-12x+9)\,\mathrm dx[/tex]
which means you can compute the integral easily with a substitution
[tex]u=x^3-6x^2+9x+3\implies\mathrm du=(3x^2-12x+9)\,\mathrm dx[/tex]
Under this transformation, the integral is
[tex]\displaystyle\int u\,\mathrm du=\frac{u^2}2+C=\boxed{\frac{(x^3-6x^2+9x+3)^2}2+C}[/tex]
On the other hand, in case you're missing a symbol and the integral is actually
[tex]\displaystyle\int\frac{x^3-6x^2+9x+3}{3x^2-12x+9}\,\mathrm dx[/tex]
then first carry out the division:
[tex]\dfrac{x^3-6x^2+9x+3}{3x^2-12x+9}=\dfrac x3-\dfrac23-\dfrac{2x-9}{3x^2-12x+9}[/tex]
Now, [tex]3x^2-12x+9=3(x-3)(x-1)[/tex], so to integrate the remainder term you can decompose it into partial fractions:
[tex]-\dfrac{2x-9}{3(x-3)(x-1)}=\dfrac a{x-3}+\dfrac b{x-1}[/tex]
[tex]9-2x=a(x-1)+b(x-3)[/tex]
[tex]x=1\implies7=-2b\implies b=-\dfrac72[/tex]
[tex]x=3\implies3=2a\implies a=\dfrac32[/tex]
[tex]\implies-\dfrac{2x-9}{3(x-3)(x-1)}=\dfrac 3{2(x-3)}-\dfrac 7{2(x-1)}[/tex]
Then the integral would be
[tex]\displaystyle\int\frac{x^3-6x^2+9x+3}{3x^2-12x+9}\,\mathrm dx=\boxed{\frac{x^2}6-\frac{2x}3+\frac32\ln|x-3|-\frac72\ln|x-1|+C}[/tex]
which can be rewritten in several ways, such as
[tex]\dfrac{x^2-4x}6+\dfrac12ln\left|\dfrac{(x-3)^3}{(x-1)^7}\right|+C[/tex]
