Answer
$3+\ln\frac{3}{8}$
Work Step by Step
$\int_1^2\frac{3x^{2}+6x+2}{x^{2}+3x+2}dx$ = $\int_1^2\left(3+\frac{-3x-4}{x^{2}+3x+2}\right)dx$ = $\int_1^2\left[3+\frac{-3x-4}{(x+2)(x+1)}\right]dx$ = $\int_1^2\left[3+\frac{A}{x+1}+\frac{B}{x+2}\right]dx$
$-3x-4$ = $A(x+2)+B(x+1)$
$-3$ = $A+B$
$-4$ = $2A+B$
$A$ = $-1$, $B$ = $-2$
So
$\int_1^2\frac{3x^{2}+6x+2}{x^{2}+3x+2}dx$ = $\int_1^2\left(3-\frac{1}{x+1}-\frac{2}{x+2}\right)dx$ = $[3x-\ln|x+1|-2\ln|x+2|]_1^2$ = $(6-\ln3-2\ln4)-(3-\ln2-2\ln3)$ = $3+\ln(\frac{3}{8})$
