<s>[INST] integral $\int \frac{2x}{(3x-2)^2+2^2} \, \mathrm{d}x$ 
$$\int \frac{2x}{(3x-2)^2+2^2}\, \mathrm{d}x$$

$\int \frac{1}{(3x-2)^2+2^2}\, \mathrm{d}x=\frac{1}{2}\arctan(\frac{3x-2}{2})+C$
But what can I do with the $2x$ in the numerator?
 [/INST] This can be written as:
$$\frac{1}{9} \int \frac{(18x-12+12)dx}{9x^2-12x+8}=$$
$$\frac{1}{9} \int \frac{(18x-12)dx}{9x^2-12x+8}+\frac{4}{3} \int \frac{dx}{(3x-2)^2+2^2}=$$
$$\frac{1}{9} \ln(9x^2-12x+8)+\frac{2}{9}\tan^{-1}{\frac{3x-2}{2}}+C$$
</s>