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Definite Integrals Part 2

Section: Mathematics

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Question : 62 of 100

Marks: +1, -0

\int\limits_{6}^{16} \frac{\log_{e} x^{2}}{\log_{e} x^{2}+\log_{e}(x^{2}-44x+484)} dx

[27 Aug 2021 Shift 1]

6
8
5
10

Solution:

Let

I=\int\limits_{6}^{16} \frac{\ln_{e}(x^{2})}{\ln_{e}(x^{2})+\ln_{e}(484-44x+x^{2})} dx

=\int\limits_{6}^{16} \frac{\ln_{e}(x^{2})}{\ln_{e}(x^{2})+\ln_{e}(22-x)^{2}} dx

=\int\limits_{6}^{16} \frac{2 \ln_{e} x dx}{2 \ln_{e} x+2 \ln_{e}(22-x)}

I=\int\limits_{6}^{16} \frac{\ln_{e} x dx}{\ln_{e} x+\ln_{e}(22-x)}

...(i)

\because \int\limits_{a}^{b} f(x) dx = \int\limits_{a}^{b} f(a+b-x) dx

\therefore I=\int\limits_{6}^{16} \frac{\ln_{e}(22-x)}{\ln_{e}(22-x)+\ln_{e} x} dx

...(ii)

Adding Eqs. (i) and (ii), we get

2I=\int\limits_{6}^{16} \frac{\ln_{e} x+\ln_{e}(22-x)}{\ln_{e} x+\ln_{e}(22-x)} dx

2I=\int\limits_{6}^{16} dx = \left. x \right|_{6}^{16} = 10

or I = 5

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