<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] $$\int \frac{x}{2x^2+x+3}dx$$
$$=\frac{1}{4} \int \frac{4x+1}{2x^2+x+3} dx- \frac{1}{4} \int \frac{1}{2x^2+x+3}$$
I presume you know how to solve the first integral [Do so using the sub: $u=2x^2+x+3$].
Complete the square of the second and substitute $m=\sqrt{2}x+\frac{1}{2\sqrt{2}}$to get:
$$\frac{-1}{4}\int\frac{1}{\left(\sqrt{2}x+\frac{1}{2\sqrt{2}}\right)^2+\frac{23}{8}}dx=\frac{-1}{4}\int\frac{1}{\sqrt{2}}\frac{1}{(m^2+\frac{23}{8})}dm$$
Factor out the 23/8 and the constants, to get $$\frac{-\sqrt{2}}{23}\int\frac{1}{\frac{8m^2}{23}+1}dm$$
Substitute $q=2\sqrt{\frac{2m}{23}}$, solve the integral (it should be easy knowing the derivative of arctan(q)). Substitute back everything.
You should eventually get the below as the answer:
$$\frac{1}{4}\ln(2x^2+x+3)-\frac{\tan^{-1}\left(\frac{4x+1}{\sqrt{23}}\right)}{2\sqrt{23}}+C$$
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