${\int }_{6}^{16}\frac{{\log }_{e}{x}^{2}}{{\log }_{e}{x}^{2}+{\log }_{e}\left({x}^{2}-44x+484\right)}dx$ is equal to [JEE Main 2021, 27 Aug (Shift 1)]

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\({\int }_{6}^{16}\frac{{\log }_{e}{x}^{2}}{{\log }_{e}{x}^{2}+{\log }_{e}\left({x}^{2}-44x+484\right)}dx\) is equal to …

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