Answer
$$\int_{1}^{2}\frac{3x^{2}+6x+2}{x^{2}+3x+2}dx=3+ln\frac{3}{8}$$
Work Step by Step
$$\int_{1}^{2}\frac{3x^{2}+6x+2}{x^{2}+3x+2}dx=\int_{1}^{2}(3+\frac{-3x-4}{x^{2}+3x+2})dx$$
$$=\int_{1}^{2}(3+\frac{A}{x+2}+\frac{B}{x+1})dx$$
$3x^{2}+(A+B+9)x+(A+2B+6)=3x^{2}+6x+2. So\ A=-2, B=-1$
$$\int_{1}^{2}\frac{3x^{2}+6x+2}{x^{2}+3x+2}dx=\int_{1}^{2}(3-\frac{2}{x+2}-\frac{1}{x+1})dx$$
$$=\left [ 3x-2ln|x+2|-ln|x+1| \right ]_{1}^{2}$$
$$=6-2ln4-ln3-3+2ln3+ln2=3+ln\frac{3}{8}$$