${\int }_6^{16}\frac{{\log }_e{x}^2}{{\log }_e{x}^2+{\log }_e\left(x^2-44x+484\right)}dx$ is equal to

Let ${\int }_6^{16}\frac{{\log }_e{x}^2 dx}{{\log }_e{x}^2+{\log }_e\left(x^2-44x+484\right)}$

${\int }_6^{16}\frac{{\log }_e{x}^2}{{\log }_e{x}^2+{\log }_e{\left(x-22\right)}^2}dx .....\left(1\right)$

We know

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$

So, ${\int }_6^{16}\frac{{\log }_e{\left(22-x\right)}^2}{{\log }_e{\left(22-x\right)}^2+{\log }_e{\left(22-\left(22-x\right)\right)}^2}dx$

$\mathrm{I}={\int }_6^{16}\frac{{\log }_{\mathrm{e}}{\left(22-\mathrm{x}\right)}^2}{{\log }_{\mathrm{e}}{\mathrm{x}}^2+{\log }_{\mathrm{e}}{\left(22-\mathrm{x}\right)}^2}\mathrm{dx} ....\left(2\right)$

Add $\left(1\right) \& \left(2\right)$

$2I={\int }_6^{16}dx=10,I=5$