$\int \dfrac{dx}{3x^2 + 13x - 10}$

The correct answer is $\dfrac{1}{17} log \left| \dfrac{3x - 2}{3x + 15} \right|$ + $C$

Consider $3x^2 + 13x - 10$ = $3 \left( x^2 + \dfrac{13}{3}x - \dfrac{10}{3} \right)
= \left[ \left( x + \dfrac{13}{6} \right)^2 - \left( \dfrac{17}{6} \right)^2 \right]$

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

= $\dfrac{1}{3} \dfrac{1}{2} \dfrac{6}{17} log \left( \dfrac{x + \dfrac{13}{6} - \dfrac{17}{6}}{x + \dfrac{13}{6} + \dfrac{17}{6}} \right)$ + $C$

= $\dfrac{1}{17} log \left( \dfrac{x - \dfrac{4}{6}}{x + \dfrac{30}{6}} \right)$ + $C$

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

= $\dfrac{1}{17} log \left| \dfrac{3x - 2}{3x + 15} \right|$ + $C$