MCQ
$\int \limits_{6}^{16} \frac{\log _{\mathrm{e}} x^{2}}{\log _{e} x^{2}+\log _{e}\left(x^{2}-44 x+484\right)} d x$ is equal to:
- A$6$
- B$8$
- ✓$5$
- D$10$
$I=\int_{6}^{16} \frac{\log _{e} x^{2}}{\log _{e} x^{2}+\log _{e}(x-22)^{2}} d x \ldots(1)$
We know
$\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x)\, d x(\text { king })$
So $I=\int_{6}^{16} \frac{\log _{e}(22-x)^{2}}{\log _{e}(22-x)^{2}+\log _{e}(22-(22-x))^{2}}$
$I=\int_{0}^{16} \frac{\log _{e}(22-x)^{2}}{\log _{e} x^{2}+\log _{e}(22-x)^{2}} \,d x \ldots(2)$
$(1)+(2)$
$2 I=\int_{6}^{16} 1 .\, d x=10$
$I=5$
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