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

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Definite Integrals

\int_{6}^{16} \frac{l o g _{e} x ^{2}}{l o g _{e} x ^{2} + l o g _{e} ( x ^{2} - 44 x + 484 )} d x

is equal to

Answer: (D)

Let

I = \int_{6}^{16} \frac{ln x ^{2}}{ln x ^{2} + ln ( x ^{2} - 44 x + 484 )} d x

, and note

x^{2} - 44 x + 484 = (x - 22)^{2}

. Under

x \to 22 - x

this quadratic becomes

x^{2}

while

x^{2}

becomes

(22 - x)^{2}

, so

I = \int_{6}^{16} \frac{ln ( 22 - x ) ^{2}}{ln ( 22 - x ) ^{2} + ln x ^{2}} d x .

Adding the two forms, the integrand sums to

1

, giving

2 I = \int_{6}^{16} d x = 10

, so

I = 5.

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