Evaluating $\int \dfrac {2x} {x^{2} + 6x + 13}dx$ I am having trouble understanding the first step of evaluating
$$\int \dfrac {2x} {x^{2} + 6x + 13}dx$$ 
When faced with integrals such as the one above, how do you know to manipulate the integral into:
$$\int \dfrac {2x+6} {x^{2} + 6x + 13}dx - 6 \int \dfrac {1} {x^{2} + 6x + 13}dx$$ 
After this first step, I am fully aware of how to complete the square and evaluate the integral, but I am having difficulties seeing the first step when faced with similar problems. Should you always look for what the $"b"$ term is in a given $ax^{2} + bx + c$ function to know what you need to manipulate the numerator with? Are there any other tips and tricks when dealing with inverse trig antiderivatives?

You want to make the substitution $u=x^2+6x+13$ . Thus, $du=2x+6$ the rest follows easily from there.
