The integral ∫ 2 4 log x 2 log x 2 + log 6 - x 2 d x is equal to

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

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

$\left[\because \underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx\right]$

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

$2I=\underset{2}{\overset{4}{\int }}1\mathit{dx}$

$2I={\left[x\right]}_2^4$

$\therefore I=1$