Evaluate the integral:

$I=\int \frac{1-3 x}{3 x^{2}+4 x+2} d x$

As we can see that there is a term of $x$ in numerator and derivative of $x^{2}$ is also $2 x$. So there is a chance that we can make substitution for $3 x^{2}+4 x+2$ and I can be reduced to a fundamental integration.

As, $\frac{\mathrm{d}}{\mathrm{dx}}\left(3 \mathrm{x}^{2}+4 \mathrm{x}+2\right)=6 \mathrm{x}+4$

$\therefore$ Let, $1-3 x=A(6 x+4)+B$

$\Rightarrow 1-3 x=6 A x+4 A+B$

On comparing both sides -

We have,

$6 A=-3 \Rightarrow A=-1 / 2$

$4 \mathrm{~A}+\mathrm{B}=1 \Rightarrow \mathrm{B}=-4 \mathrm{~A}+1=3$

Hence,

$I=\int \frac{-\frac{1}{2}(6 x+4)+3}{3 x^{2}+4 x+2} d x$

$\therefore I=-\frac{1}{2} \int \frac{6 x+4}{3 x^{2}+4 x+2} d x+\int \frac{3}{3 x^{2}+4 x+2} d x$

Let, $I_{1}=-\frac{1}{2} \int \frac{6 x+4}{3 x^{2}+4 x+2} d x$ and $I_{2}=\int \frac{3}{3 x^{2}+4 x+2} d x$

Now, $I=I_{1}+I_{2} \ldots$ eqn 1

We will solve $I_{1}$ and $I_{2}$ individually.

As $I_{1}=-\frac{1}{2} \int \frac{6 x+4}{3 x^{2}+4 x+2} d x$

Let $u=3 x^{2}+4 x+2 \Rightarrow d u=(6 x+4) d x$

$\therefore I_{1}$ reduces to $-\frac{1}{2} \int \frac{\text { du }}{u}$

Hence,

$\mathrm{I}_{1}=-\frac{1}{2} \int \frac{\mathrm{du}}{\mathrm{u}}=-\frac{1}{2} \log |\mathrm{u}|+\mathrm{C}\left\{\because \int \frac{\mathrm{dx}}{\mathrm{x}}=\log |\mathrm{x}|+\mathrm{C}\right\}$

On substituting the value of $u$, we have:

$\mathrm{I}_{1}=-\frac{1}{2} \log \left|3 \mathrm{x}^{2}+4 \mathrm{x}+2\right|+\mathrm{C} \ldots .$ eqn 2

As, $I_{2}=\int \frac{3}{3 x^{2}+4 x+2} d x$ and we don't have any derivative of function present in denominator. $\therefore$ we will use some special integrals to solve the problem.

As denominator doesn't have any square root term. So one of the following two integrals will solve the problem.

i) $\int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C$ ii) $\int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C$

Now we have to reduce $I_{2}$ such that it matches with any of above two forms.

We will make to create a complete square so that no individual term of $x$ is seen in denominator.

$\therefore I_{2}=\int \frac{3}{3\left(x^{2}+\frac{4}{3} x+\frac{2}{3}\right)} d x=\int \frac{1}{x^{2}+\frac{4}{3} x+\frac{2}{3}} d x$

$\Rightarrow I_{2}=\int \frac{1}{\left\{x^{2}+2\left(\frac{2}{3}\right) x+\left(\frac{2}{3}\right)^{2}\right\}+\frac{2}{3}-\left(\frac{2}{3}\right)^{2}} d x$

Using: $a^{2}+2 a b+b^{2}=(a+b)^{2}$

We have:

$\mathrm{I}_{2}=\int \frac{1}{\left(\mathrm{x}+\frac{2}{3}\right)^{2}+\left(\frac{\sqrt{2}}{3}\right)^{2}} \mathrm{dx}$

$\mathrm{I}_{2}$ matches with $\int \frac{1}{\mathrm{x}^{2}+\mathrm{a}^{2}} \mathrm{dx}=\frac{1}{\mathrm{a}} \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{C}$

$\therefore I_{2}=\frac{1}{\frac{\sqrt{2}}{3}} \tan ^{-1}\left(\frac{x+\frac{2}{2}}{\frac{\sqrt{2}}{3}}\right)+C$

$\therefore I_{2}=\frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+2}{\sqrt{2}}\right)+C \ldots$ eqn 3

From eqn 1:

$I=I_{1}+I_{2}$

Using eqn 2 and eqn 3:

$\therefore\left|=-\frac{1}{2} \log \right| 3 x^{2}+4 x+2 \mid+\frac{3}{\sqrt{2}} \tan ^{-1}\left(\frac{3 x+2}{\sqrt{2}}\right)+C$