Step-by-Step Solution
Key Concept: General
$\int \frac{dx}{3x^2+6x+15} = \frac{1}{3} \int \frac{dx}{x^2+2x+5} = \frac{1}{3} \int \frac{dx}{(x+1)^2+4} = \frac{1}{3} \int \frac{dx}{(x+1)^2+(2)^2} = \frac{1}{6} \tan^{-1} \left( \frac{x+1}{2} \right) + C$
Correct Answer: $\frac{1}{6} \tan^{-1} \left( \frac{x+1}{2} \right) + C$