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?

I look at that fraction and see that the numerator differs from the derivative of the denominator by a constant, $6$. If the numerator were $2x+6$ instead of $2x$, the fraction would be of the form $u'/u$, and I’d be very happy. So I simply make it $2x+6$, subtracting $6$ to compensate:
$$\frac{2x}{x^2+6x+13}=\frac{(2x+6)-6}{x^2+6x+13}=\frac{2x+6}{x^2+6x+13}-\frac6{x^2+6x+13}\;.$$
Then I consider whether I can integrate the correction term. In this case I recognize it as the derivative of an arctangent, so I know that I’ll be able to handle it, though it will take a little algebra.
