QJEEM15Ma8bd1: JEE Main 2015 Mathematics - Class 12 JEEM15Ma8bd1 Solution: The integral is equal to :

Question

Verified Solution • 2025–26 Session moderate

The integral

\int\limits_2^4 {{{\log \,{x^2}} \over {\log {x^2} + \log \left( {36 - 12x + {x^2}} \right)}}dx}

is equal to :

Accepted Answer

I = \int\limits_2^4 {{{\log {x^2}} \over {\log {x^2} + \log \left( {36 - 12x + {x^2}} \right)}}}

I = \int\limits_2^4 {{{\log {x^2}} \over {\log {x^2} + \log {{\left( {6 - x} \right)}^2}}}} \,\,\,\,\,\,\,\,\,\,...\left( i \right)

I = \int\limits_2^4 {{{\log {{\left( {6 - x} \right)}^2}} \over {\log {{\left( {6 - x} \right)}^2} + \log {x^2}}}} \,\,\,\,\,\,\,\,...\left( {ii} \right)

Adding

(1)

and

(2)

2I = \int\limits_2^4 {dx = \left[ x \right]_2^4} = 2 \Rightarrow 1 - 1

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