You will be trapped with this Integral

The integrand can be rewritten as

$\dfrac{6x^{2} - 5x + 1 + 4x - 2}{6x^{2} - 5x + 1} = 1 + \dfrac{2(2x - 1)}{(2x - 1)(3x - 1)} = 1 + \dfrac{2}{3x - 1}$

for $x \ne \frac{1}{2}$ . The point of discontinuity $(\frac{1}{2}, 5)$ does not affect the integral, but what is of concern is when $3x - 1 = 0$ , as this leads to the vertical asymptote $x = \frac{1}{3}$ . So splitting the integral into two improper integrals, one from $0$ to $\frac{1}{3}$ and the other from $\frac{1}{3}$ to $1$ , we end up needing to evaluate

$(x + \frac{2}{3}\ln |3x - 1|)_{0}^{\frac{1}{3}^{-}} + (x + \frac{2}{3}\ln |3x - 1|)_{\frac{1}{3}^{+}}^{1} =$

$\displaystyle \dfrac{1}{3} + \dfrac{2}{3}\lim_{x \to \frac{1}{3}^{-}} \ln(1 - 3x) + 1 + \dfrac{2}{3}\ln(2) - \dfrac{1}{3} - \dfrac{2}{3}\lim_{x \to \frac{1}{3}^{+}} \ln(3x - 1) =$

$\displaystyle 1 + \dfrac{2}{3}\ln(2) + \dfrac{2}{3}\left(\lim_{x \to \frac{1}{3}^{-}} \ln(1 - 3x) - \lim_{x \to \frac{1}{3}^{+}} \ln(3x - 1)\right)$ .

Now the bracketed pair of limits are of the indeterminate form $-\infty - (-\infty) \Longrightarrow \infty - \infty$ , and so the integral cannot be evaluated, i.e., it is $\boxed{divergent}$ .