Rewriting the given equation,
$$
\Rightarrow-\int \frac{3 x-1}{\left(3 x^{2}+4 x+2\right)} d x
$$
Using partial fractions,
$$
\begin{aligned}
&(3 x-1)-A\left(\frac{d}{d x}\left(3 x^{2}+4 x+2\right)\right)+B \\
&3 x-1=A(6 x+4)+B
\end{aligned}
$$
Equating the coefficients of \(x\),
$$
\begin{aligned}
&3=6 \mathrm{~A} \\
&\mathrm{~A}=\frac{1}{2} \\
&\text { Also, }-1=4 \mathrm{~A}+\mathrm{B} \\
&=\mathrm{B}=-3
\end{aligned}
$$
Substituting in the original equation,
$$
\begin{aligned}
&\Rightarrow-\int \frac{\frac{1}{2}(6 x+4)-3}{\left(3 x^{2}+4 x+2\right)} d x \\
&\Rightarrow-\frac{1}{2} \log \left|3 x^{2}+4 x+2\right|+3 \int \frac{1}{3\left(x^{2}+\frac{4}{3} x+\frac{2}{3}\right)} d x \\
&\text { Let } 1=3 \int \frac{1}{2\left(x^{2}+\frac{\pi}{3} x^{2}\right)} d x \\
&\Rightarrow \int \frac{1}{\left(x^{2}+\frac{4}{3} x+\frac{2}{3}\right)} d x \\
&\rightarrow \int \frac{1}{\left(\left(x+\frac{2}{3}\right)^{2}+\frac{2}{3}-\frac{4}{9}\right)} d x \\
&\rightarrow \int \frac{1}{\left(\left(x+\frac{2}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}\right)} d x \\
&\text { Here } x=\frac{\sqrt{2}}{3} \\
&\rightarrow \frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{x+\frac{2}{3}}{\frac{\sqrt{2}}{3}}\right)+c \\
&\Rightarrow \frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+2}{\sqrt{2}}\right)+c
\end{aligned}
$$
Substituting in \((1)\) and combining with ariginal equation,
$$
\rightarrow-\frac{1}{2} \log \left|3 x^{2}+4 x+2\right|+\frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+2}{\sqrt{2}}\right)+C
$$
Therefore,
$$
\int \frac{1-3 x}{\left(3 x^{2}+4 x+2\right)} d x=-\frac{1}{2} \log \left|3 x^{2}+4 x+2\right|+\frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+2}{\sqrt{2}}\right)+c
$$
