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?
 A: 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).
A: $$\frac{2x}{x^2+6x+3}=\frac{A}{x-(-3-\sqrt{6})}+\frac{B}{x-(\sqrt{6}-3)}$$
Find $A$ and $B$ and then:
$$\int \frac{2x}{x^2+6x+3} dx=\int \frac{A}{x-(-3-\sqrt{6})} dx+\int \frac{B}{x-(\sqrt{6}-3)} dx$$
A: Another idea (just reducing it to another form):
Let $$I=6\int \frac{1}{x^2+6x+3} dx=6\int \frac{1}{(x+3)^2-6} dx=\int \frac{1}{(\frac{1}{\sqrt{6}}(x+3))^2-1} dx$$.
Now let $$\frac{1}{\sqrt{6}}(x+3)=\cosh a$$, hence using $$\cosh^2 a - 1 = \sinh ^2 a$$ and $$\frac{1}{\sqrt{6}} = \sinh a \frac{da}{dx}\Leftrightarrow dx = da \sinh a \sqrt{6}$$ we get $$I=\int \frac{1}{\sinh ^2 a} da \sinh a\sqrt{6} = \sqrt{6} \int \frac{1}{\sinh a} da$$.
EDIT: Can someone please show me how to write bigger LaTeX?
EDIT2: Neat!
A: HINT:
As
$\displaystyle x^2+6x+3=(x+3)^2-(\sqrt6)^2,$
using Trigonometric substitution, set $x+3=\sqrt6\sec\theta$
or use $\#1$ of this
