Find the integral: $\displaystyle \int \frac{dx}{3x^2 + 13x - 10}$

$\frac{1}{17} \ln \left| \frac{3x - 2}{x + 5} \right| + C$

The correct answer is Option (3) → $\frac{1}{17} \ln \left| \frac{3x - 2}{x + 5} \right| + C$

We write the denominator of the integrand,

$3x^2 + 13x - 10 = 3 \left( x^2 + \frac{13x}{3} - \frac{10}{3} \right) = 3 \left[ \left( x + \frac{13}{6} \right)^2 - \left( \frac{17}{6} \right)^2 \right] \text{ (completing the square)}$$

Thus $\int \frac{dx}{3x^2 + 13x - 10} = \frac{1}{3} \int \frac{dx}{\left( x + \frac{13}{6} \right)^2 - \left( \frac{17}{6} \right)^2}$

Put $x + \frac{13}{6} = t$. Then $dx = dt$.

Therefore, $\int \frac{dx}{3x^2 + 13x - 10} = \frac{1}{3} \int \frac{dt}{t^2 - \left( \frac{17}{6} \right)^2}$

$= \frac{1}{3 \times 2 \times \frac{17}{6}} \log \left| \frac{t - \frac{17}{6}}{t + \frac{17}{6}} \right| + C_1$

$= \frac{1}{17} \log \left| \frac{x + \frac{13}{6} - \frac{17}{6}}{x + \frac{13}{6} + \frac{17}{6}} \right| + C_1$

$= \frac{1}{17} \log \left| \frac{6x - 4}{6x + 30} \right| + C_1$

$= \frac{1}{17} \log \left| \frac{3x - 2}{x + 5} \right| + C_1 + \frac{1}{17} \log \frac{1}{3}$

$= \frac{1}{17} \log \left| \frac{3x - 2}{x + 5} \right| + C, \text{ where } C = C_1 + \frac{1}{17} \log \frac{1}{3}$