Completing the square:

$$\int\frac{5x+6}{x^2+3x+1}dx=\int\frac{5x+6}{\left(x+\frac{3}{2}\right)^2-\frac{5}{4}}dx$$

Letting $t=x+\frac{3}{2}$ leads to

$$\begin{align*}&\int\frac{5t-\frac{3}{2}}{t^2-\frac{5}{4}}dt\\&=5\int\frac{t}{t^2-\frac{5}{4}}dt-\frac{3}{2}\int\frac{dt}{t^2-\frac{5}{4}}\\&=\frac{5}{2}\ln\left|t^2-\frac{5}{4}\right|-\frac{3}{2\sqrt{5}}\ln\left|\frac{2t-\sqrt{5}}{2t+\sqrt{5}}\right|+C\\&=\boxed{\frac{5}{2}\ln\left|(x^2+3x+1)\right|-\frac{3}{2\sqrt{5}}\ln\left|\frac{2x+3-\sqrt{5}}{2x+3+\sqrt{5}}\right|+C}\end{align*}$$