Question

The integral ${\int }_2^4\frac{\log x^2}{\log x^2+\log (36-12x+x^2)}dx$ is equal to :-

• A. $2$
• B. $4$
• C. $1$
• D. $6$

Answer 1

The correct option is C $1$

$\begin{array}{c}Wehave\\ I={\int }_2^4\frac{\log x^2}{\log x^2+\log {\left(6-x\right)}^{2}dx}\\ ={\int }_2^4\frac{\log x}{\log x+\log \left(6-x\right)}dx----\left(1\right)\\ u\sin gformulaf\left(a+b-x\right)=f\left(x\right)\\ I={\int }_2^4\frac{\log \left(6-x\right)}{\log \left(6-x\right)+\log x}dx----\left(2\right)\\ Nowonadding\left(1\right)+\left(2\right)weget\\ 2I={\int }_2^4\frac{\log \left(6-x\right)}{\log \left(6-x\right)+\log x}dx\\ 2I={\int }_2^{4}dx\\ 2I=4-2\\ 2I=2\\ \therefore I=1\\ Hence,optionCisthecorrectanswer.\end{array}$