Evaluate the definite integral $\int_{1}^{2} \frac{5 x^{2}}{x^{2}+4 x+3} dx$.

(A) Let $I = \int_{1}^{2} \frac{5 x^{2}}{x^{2}+4 x+3} dx$.
Dividing $5x^2$ by $x^2+4x+3$,we get $5 - \frac{20x+15}{x^2+4x+3}$.
Thus,$I = \int_{1}^{2} 5 dx - \int_{1}^{2} \frac{20x+15}{x^2+4x+3} dx = [5x]_{1}^{2} - I_1 = 5 - I_1$,where $I_1 = \int_{1}^{2} \frac{20x+15}{x^2+4x+3} dx$.
To solve $I_1$,let $20x+15 = A(2x+4) + B$. Comparing coefficients,$2A = 20 \Rightarrow A = 10$ and $4A+B = 15 \Rightarrow 40+B = 15 \Rightarrow B = -25$.
So,$I_1 = \int_{1}^{2} \frac{10(2x+4) - 25}{x^2+4x+3} dx = 10 \int_{1}^{2} \frac{2x+4}{x^2+4x+3} dx - 25 \int_{1}^{2} \frac{dx}{(x+2)^2 - 1^2}$.
$I_1 = [10 \ln|x^2+4x+3|]_{1}^{2} - 25 [\frac{1}{2} \ln|\frac{x+2-1}{x+2+1}|]_{1}^{2} = [10 \ln|x^2+4x+3|]_{1}^{2} - \frac{25}{2} [\ln|\frac{x+1}{x+3}|]_{1}^{2}$.
Evaluating the limits: $I_1 = (10 \ln 15 - 10 \ln 8) - \frac{25}{2} (\ln \frac{3}{5} - \ln \frac{2}{4}) = 10 \ln \frac{15}{8} - \frac{25}{2} \ln \frac{6}{5} = 10 \ln \frac{15}{8} - \frac{25}{2} \ln \frac{6}{5}$.
Finally,$I = 5 - (10 \ln \frac{15}{8} - \frac{25}{2} \ln \frac{6}{5}) = 5 - 10 \ln \frac{15}{8} + \frac{25}{2} \ln \frac{6}{5}$.