<s>[INST] Compute $\int\frac{x}{2x^2+x+3}dx$ $$\int\frac{x}{2x^2+x+3}\,dx$$
Well, to approach this kind of exercises I know that I need to check the derivative of the denominator. which is $4x + 1$.
Then, I can re-write the integral: $\int\frac{0.25(4x+1) - 0.25}{2x^2+x+3}\,dx$.
Then, I get:
$$0.25\int\frac{(4x+1)}{2x^2+x+3}\,dx-0.25\int\frac{1}{2x^2+x+3}\,dx.$$
The left one is pretty straight forward with $\ln|\cdot|$, 
Problem: does anyone have some "technique" to solve the right integral? hints would be appreciated too.
Edit: maybe somehow: $$0.25\int\frac{1}{2(2x^2/2+x/2+3/2)}\,dx = 0.25\int\frac{1}{(x+0.25)^2 + \frac{23}{16}}\,dx$$
 [/INST] When you have a second degree polynomial in the denominator with $\Delta <0$, you can proceed like this: find the solutions, in this case $\frac{-1}{4} \pm \frac{i\sqrt {23}}{4}$, and then write the polynomial in the usual way
$$2\Big(x+\frac{1}{4} + \frac{i\sqrt {23}}{4}\Big)\Big(x+\frac{1}{4} - \frac{i\sqrt {23}}{4}\Big) $$
then you can use the fact that $(a-b)(a+b)=a^2-b^2$ to deduce
$$2x^2+x+3=2\Big(x+\frac{1}{4} \Big)^2+\frac{23}{8}.$$
Then apply a substitution to get $\frac{1}{1+y^2}$.
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