= Solution
Let $D=3x^2+2x+2$. Splitting the numerator into a <derivative> term and a constant gives $x-1=D'/6-4/3$. Thus
$$
\int\frac{x-1}{D}\,dx=\frac16\log D-\frac43\int\frac{dx}{D}.
$$
Complete the square as $D=((3x+1)^2+5)/3$ and substitute $u=3x+1$. The remaining <integral> is $\int du/(u^2+5)=\arctan(u/\sqrt5)/\sqrt5$. Hence
$$
\boxed{\frac16\log(3x^2+2x+2)
-\frac4{3\sqrt5}\arctan\frac{3x+1}{\sqrt5}+C}.
$$
The quadratic denominator is positive for all real $x$, so no logarithmic absolute value or real-domain exclusion is needed.
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