STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

$\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx$ and then add

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

$I = \int\limits_2^4 {\frac{{2\log xdx}}{{2\log {x^2} + \log {{\left( {6 - x} \right)}^2}}}} dx$

$\int\limits_2^4 {\frac{{2\log x}}{{2[\log x + \log (6 - x)]}}} $

$ \Rightarrow I = \int\limits_2^4 {\frac{{\log xdx}}{{\left| {\log x + \log \left( {6 - x} \right)} \right|}}} ...........\left( 1 \right)$

$ \Rightarrow I = \int\limits_2^4 {\frac{{\log \left( {6 - x} \right)}}{{\log \left( {6 - x} \right) + \log x}}dx} ...........\left( 2 \right)$

$\left[ {\therefore \int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx} \right]$

On adding Eqs. $(i)$ and $(ii),$ we get

$2I = \int\limits_2^4 {\frac{{\log x + \log \left( {6 - x} \right)}}{{\log x + \log \left( {6 - x} \right)}}dx} $

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

$ \Rightarrow 2I = 2$

$\Rightarrow \quad 2 I=2 \Rightarrow I=1$