<s>[INST] Integrate a division of polynomials Hi I have the following integral:
$$\int \frac{2x}{x^2+6x+3}\, dx$$
I made some changes like:
$$\int \dfrac{2x+6-6}{x^2+6x+3}\, dx$$
then I have: 
$$\int \dfrac{2x+6}{x^2+6x+3}\, dx -\int\dfrac{6}{x^2+6x+3}\, dx$$
and thus: $$\ln(x^2+6x+3)-\int\dfrac{6}{x^2+6x+3}\, dx$$
Ok, I have decomposed $$\frac{2x}{x^2+6x+3} $$  in: $$ \frac{3+\sqrt6}{\sqrt6(x+\sqrt 6+3)} + \frac{3-\sqrt6}{\sqrt6 (-x+\sqrt6-3)}$$
How can I integrate this expressions?
 [/INST] A start: Note that $x^2+6x+3=0$ has the roots $\alpha=-3+\sqrt{6}$ and $\beta=-3-\sqrt{6}$. Thus $x^2+6x+3=(x-\alpha)(x-\beta)$.
Express $\frac{6}{(x-\alpha)(x-\beta)}$ as $\frac{A}{x-\alpha}+\frac{B}{(x-\beta)}$ (partial fractions).
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