Find the derivative of $$g(x) = \int_{3x}^{6x} \frac{u+1}{u^2+4} du$$
$g'(x) =$
HINT: $$\int_{3x}^{6x} \frac{u+1}{u^2+4} du = \int_{0}^{6x} \frac{u+1}{u^2+4} du + \int_{3x}^{0} \frac{u+1}{u^2+4} du$$