# How do you integrate x/((ln3)(x^2 + 4))?

$\int \frac{x}{\left(\ln 3\right) \left({x}^{2} + 4\right)} \mathrm{dx}$
$\ln 3$ is a constant so this turns down to,
$\implies \frac{1}{\ln} 3 \int \frac{x}{{x}^{2} + 4} \mathrm{dx} = \frac{1}{\ln} 3 \int \frac{1}{2} \cdot \frac{2 x}{{x}^{2} + 4} \mathrm{dx} = \frac{1}{2 \ln 3} \int \frac{2 x}{{x}^{2} + 4} \mathrm{dx}$
$\implies \frac{1}{2 \ln 3} \ln \left\mid {x}^{2} + 4 \right\mid + C$