$$\int\frac{5x+6}{(x^2+x+2)^2}dx=\int\frac{5x+6}{\left[\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\right]^2}dx$$

Let $t=x+\frac{1}{2}$:

$$\int\frac{5t+\frac{7}{2}}{\left(t^2+\frac{7}{4}\right)^2}dt=5\int\frac{t}{\left(t^2+\frac{7}{4}\right)^2}dt+\frac{7}{2}\int\frac{dt}{\left(t^2+\frac{7}{4}\right)^2}$$

The first integral can be computed as

$$\int\frac{t}{\left(t^2+\frac{7}{4}\right)^2}dt=\frac{1}{2}\int\frac{d\left(t^2+\frac{7}{4}\right)}{\left(t^2+\frac{7}{4}\right)^2}=-\frac{1}{2}\frac{1}{t^2+\frac{7}{4}}+C$$

The second integral can be evaluated using trigonometric substitution as

$$\int\frac{dt}{\left(t^2+\frac{7}{4}\right)^2}=\frac{2}{7}\frac{t}{t^2+\frac{7}{4}}+\frac{4}{7\sqrt{7}}\arctan\frac{2t}{7}+C$$

Setting these two back and recovering $x$ from $t$ give the final result as 

$$\boxed{\frac{2x-4}{2(x^2+x+2)}+\frac{2}{\sqrt{7}}\arctan{\frac{2x+1}{\sqrt{7}}}+C}$$