${\int }_{2-\log 3}^{3+\log 3} \frac{\log {\displaystyle }(4+x)}{\log {\displaystyle }(4+x)+\log (9-x)}dx$

  I=${\int }_{2-\log 3}^{3+\log 3} \frac{\log {\displaystyle }(4+x)}{\log {\displaystyle }(4+x)+\log (9-x)}dx$

    Apply the formula ${\int }_a^b f\left(x\right)dx={\int }_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x})\mathrm{dx}$ 

$\begin{array}{l}I={\int }_{2-\log 3}^{3+\log 3} \frac{\log (9-x)}{\log (9-x)+\log (4+x)}dx\\ 2I={\int }_{2-\log 3}^{3+\log 3} dx=[x{]}_{2-\log 3}^{3+\log 3}\\ =1+2\log 3\Rightarrow I=\frac{1}{2}+\log 3\end{array}$