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: $$0.25\int\frac{1}{2(2x^2/2+x/2+3/2)}\,dx = 0.25\int\frac{1}{(x+0.25)^2 + \frac{23}{16}}\,dx$$

$$\int \frac{1}{2x^2+x+3}\, dx=\frac{1}{\sqrt{2}}\int \frac{d\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)}{\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)^2+\frac{23}{8}}\, $$
Let $\sqrt{2}x+\frac{1}{2\sqrt{2}}=\sqrt{\frac{23}{8}}\tan u$.
$$=\frac{1}{\sqrt{2}}\int \frac{\sqrt{\frac{23}{8}}\frac{1}{\cos^2 u}}{\frac{23}{8}(\tan^2 u+1)}\, du=\frac{2}{\sqrt{23}}\int  du=\frac{2}{\sqrt{23}}u+C$$
