Manipulate the numerator to match the derivative of the denominator.

Let the denominator be $f(x) = 3x^2+4x+2$. Its derivative is $f'(x) = 6x+4$.
We want to express the numerator $2x$ in terms of $6x+4$.
We can write $2x = A(6x+4) + B$.
Comparing coefficients:
For $x$: $2 = 6A \implies A = \frac{1}{3}$.
For constants: $0 = 4A + B \implies 0 = 4\left(\frac{1}{3}\right) + B \implies B = -\frac{4}{3}$.
So, $2x = \frac{1}{3}(6x+4) - \frac{4}{3}$.

For the first integral, let $u = 3x^2+4x+2$. Then $du = (6x+4) dx$.

(Note: $3x^2+4x+2$ is always positive because its discriminant $4^2 - 4(3)(2) = 16-24 = -8 < 0$ and the leading coefficient is positive.)

For the second integral, consider the denominator $3x^2+4x+2$.

Complete the square for $x^2 + \frac{4}{3}x$:

So, $3x^2+4x+2 = 3\left[\left(x + \frac{2}{3}\right)^2 + \frac{2}{9}\right]$.
The second integral becomes:

This integral is of the form $\int \frac{1}{u^2+a^2} du = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C$.
Here, $u = x + \frac{2}{3}$ (so $du = dx$) and $a = \frac{\sqrt{2}}{3}$.

Rationalize the denominator:

The final answer is $\boxed{\frac{1}{3} \ln(3x^2+4x+2) - \frac{2\sqrt{2}}{3} \arctan\left(\frac{3x+2}{\sqrt{2}}\right) + C}$.