Question:
$\int_{6}^{16} \frac{\log _{e} x^{2}}{\log _{e} x^{2}+\log _{e}\left(x^{2}-44 x+484\right)} d x$ is equal to:
Correct Option: , 3
Solution:
Let $I=\int_{6}^{16} \frac{\log _{e} x^{2}}{\log _{e} x^{2}+\log _{e}\left(x^{2}-44 x+484\right)} d x$
$I=\int_{6}^{16} \frac{\log _{e} x^{2}}{\log _{e} x^{2}+\log _{e}(x-22)^{2}} d x$..(1)
We know
$\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$ (king)
So $\mathrm{I}=\int_{6}^{16} \frac{\log _{\mathrm{c}}(22-\mathrm{x})^{2}}{\log _{\mathrm{e}}(22-\mathrm{x})^{2}+\log _{\mathrm{e}}(22-(22-\mathrm{x}))^{2}}$
$I=\int_{0}^{16} \frac{\log _{e}(22-x)^{2}}{\log _{e} x^{2}+\log _{e}(22-x)^{2}} d x$..(2)
$(1)+(2)$
$2 \mathrm{I}=\int_{6}^{16} 1 . \mathrm{dx}=10$
$I=5$