int limits_{6}^{16} frac{log _{e} x^{2}}{log | vedclass

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(C) Let $I = \int_{6}^{16} \frac{\log _{e} x^{2}}{\log _{e} x^{2} + \log _{e}(x-22)^{2}} dx \dots(1)$Using the property $\int_{a}^{b} f(x) dx = \int_{

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$5$

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(C) Let $I = \int_{6}^{16} \frac{\log _{e} x^{2}}{\log _{e} x^{2} + \log _{e}(x-22)^{2}} dx \dots(1)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = 6+16 = 22$:$I = \int_{6}^{16} \frac{\log _{e}(22-x)^{2}}{\log _{e}(22-x)^{2} + \log _{e}(22-(22-x))^{2}} dx$$I = \int_{6}^{16} \frac{\log _{e}(22-x)^{2}}{\log _{e}(22-x)^{2} + \log _{e} x^{2}} dx \dots(2)$Adding $(1)$ and $(2)$:$2I = \int_{6}^{16} \frac{\log _{e} x^{2} + \log _{e}(22-x)^{2}}{\log _{e} x^{2} + \log _{e}(22-x)^{2}} dx$$2I = \int_{6}^{16} 1 dx = [x]_{6}^{16} = 16 - 6 = 10$$I = \frac{10}{2} = 5$

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

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