Thanks,
$\displaystyle \int_{0}^{12} \frac{\log_{4} (4+x)}{4+x}\cdot dx = \frac{1}{\ln 4}\cdot \int_{0}^{12} \frac{\ln (4+x)}{4+x}\cdot dx =$
$\displaystyle = \frac{1}{2\cdot \ln 4}\cdot |\ln^{2}(4+x)|_{0}^{12} = \frac{3}{2}\cdot \ln 4$
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$