QJEEM21Ma0f7a: JEE Main 2021 Mathematics - Class 12 JEEM21Ma0f7a Solution

Question

Verified Solution • 2025–26 Session moderate

\int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 44x + 484)}}dx}

is equal to :

Accepted Answer

Let

I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 44x + 484)}}dx}

I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 22)}}dx}

.... (1)

We know,

\int\limits_a^b {f(x)dx} = \int\limits_a^b {f(a + b - x)} \,dx

(king)

So,

I = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{{(22 - x)}^2} + {{\log }_e}{{(21 - (22 - x))}^2}}}}

I = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{x^2} + {{\log }_e}{{(22 - x)}^2}}}dx}

.... (2)

(1) + (2)

2I = \int\limits_6^{16} {1.\,dx} = 10

I = 5

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