We need the domain of $f(x)=\log_e\left(\frac{6x^2+5x+1}{2x-1}\right)+\cos^{-1}\left(\frac{2x^2-3x+4}{3x-5}\right).$

For the domain:

1. The logarithm requires $\frac{6x^2+5x+1}{2x-1}>0.$
2. The inverse cosine requires $-1\le \frac{2x^2-3x+4}{3x-5}\le 1,$ and also $3x-5\ne 0$.

We find both conditions and intersect them.

1. Condition from the logarithm

We factor the numerator: $6x^2+5x+1=(3x+1)(2x+1).$ So we need $\frac{(3x+1)(2x+1)}{2x-1}>0.$ Critical points are $x=-\frac13,\,-\frac12,\,\frac12.$

Arrange them: $-\frac12< -\frac13< \frac12.$

Now check signs interval-wise:

• For $x< -\frac12$: expression is negative.
• For $-\frac12<x< -\frac13$: expression is positive.
• For $-\frac13<x< \frac12$: expression is negative.
• For $x>\frac12$: expression is positive.

Hence logarithm domain is $\left(-\frac12,-\frac13\right)\cup\left(\frac12,\infty\right).$

2. Condition from $\cos^{-1}$

We need $-1\le \frac{2x^2-3x+4}{3x-5}\le 1.$

We solve the two inequalities separately.

(i) Solve

$\frac{2x^2-3x+4}{3x-5}\le 1$

Bring all terms to one side: $\frac{2x^2-3x+4-(3x-5)}{3x-5}\le 0$ $\frac{2x^2-6x+9}{3x-5}\le 0.$

Now $2x^2-6x+9=2\left(x-\frac32\right)^2+\frac92>0 \quad \text{for all }x.$ So the numerator is always positive. Therefore the fraction is $\le 0$ only when the denominator is negative: $3x-5<0 \implies x<\frac53.$

Thus, $\frac{2x^2-3x+4}{3x-5}\le 1 \iff x<\frac53.$

(ii) Solve

$\frac{2x^2-3x+4}{3x-5}\ge -1$

Bring all terms to one side: $\frac{2x^2-3x+4+(3x-5)}{3x-5}\ge 0$ $\frac{2x^2-1}{3x-5}\ge 0.$

Factor numerator roots: $2x^2-1=0 \implies x=\pm \frac{1}{\sqrt2}.$ Critical points are $-\frac1{\sqrt2},\ \frac1{\sqrt2},\ \frac53.$

Sign analysis for $\frac{2x^2-1}{3x-5}:$

• $x< -\frac1{\sqrt2}$: numerator positive, denominator negative $\Rightarrow$ negative.
• $-\frac1{\sqrt2}\le x\le \frac1{\sqrt2}$: numerator non-positive, denominator negative $\Rightarrow$ non-negative.
• $\frac1{\sqrt2}<x<\frac53$: numerator positive, denominator negative $\Rightarrow$ negative.
• $x>\frac53$: numerator positive, denominator positive $\Rightarrow$ positive.

Hence, $\frac{2x^2-3x+4}{3x-5}\ge -1 \iff \left[-\frac1{\sqrt2},\frac1{\sqrt2}\right]\cup\left(\frac53,\infty\right).$

Combine (i) and (ii)

We need both, so intersect: $x<\frac53$ and $x\in \left[-\frac1{\sqrt2},\frac1{\sqrt2}\right]\cup\left(\frac53,\infty\right).$ Thus the $\cos^{-1}$ condition becomes $x\in \left[-\frac1{\sqrt2},\frac1{\sqrt2}\right].$

3. Intersect both domain conditions

Log condition: $\left(-\frac12,-\frac13\right)\cup\left(\frac12,\infty\right).$

$\cos^{-1}$ condition: $\left[-\frac1{\sqrt2},\frac1{\sqrt2}\right].$

Intersection: $\left(-\frac12,-\frac13\right)\cup\left(\frac12,\frac1{\sqrt2}\right].$

So, $(\alpha,\beta)\cup(\gamma,\delta]=\left(-\frac12,-\frac13\right)\cup\left(\frac12,\frac1{\sqrt2}\right].$

Therefore, $\alpha=-\frac12,\quad \beta=-\frac13,\quad \gamma=\frac12,\quad \delta=\frac1{\sqrt2}.$

4. Compute the required value

We need $18\left(\alpha^2+\beta^2+\gamma^2+\delta^2\right).$

Now, $\alpha^2=\frac14,\quad \beta^2=\frac19,\quad \gamma^2=\frac14,\quad \delta^2=\frac12.$

Sum: $\frac14+\frac19+\frac14+\frac12=\frac12+\frac12+\frac19=1+\frac19=\frac{10}{9}.$

Hence, $18\left(\frac{10}{9}\right)=20.$

Final Answer

$\boxed{20}$

The derived answer matches the stored correct answer.