The antiderivative of 1/(x² - 9) with C = 0 has the form F(x) = c ln|(x - 3)/(x + 3)| for x > 3…

Step 1. The standard form. With $a = 3$, $\displaystyle\int \dfrac{dx}{x^2 - 9} = \dfrac{1}{6}\ln\left|\dfrac{x - 3}{x + 3}\right| + C$. So $c = \dfrac{1}{6} = 0.1667$.

Step 2. Evaluate. At $x = 5$, $\dfrac{x - 3}{x + 3} = \dfrac{2}{8} = \dfrac{1}{4}$. So $F(5) = \dfrac{1}{6}\ln\dfrac{1}{4} = -\dfrac{\ln 4}{6} = -0.231$.