Discussion Forum

 The  Integral $\int_{2}^{4} \frac{\log x^{2}}{\log x^{2}+\log (36-12x+x^{2})}dx$    is equal to

Answer:

Explanation:

Central Idea.

   Apply  the property    $\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx$  and then add.

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

                    $I= \int_{2}^{4} \frac{2\log x}{2\log x+\log(6-x)^{2}}dx$

                        $= \int_{2}^{4} \frac{2\log x dx}{2[\log x+\log(6-x)^{}]}$

                       $\Rightarrow I=\int_{2}^{4} \frac{\log x dx}{[\log x+\log(6-x)^{}]}$      ......(i)

        $\Rightarrow I=\int_{2}^{4} \frac{\log (6-x) }{\log (6-x)+\log x}dx$    .......(ii)

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

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

                 $2I= \int_{2}^{4}\frac{\log x +\log (6-x)}{\log x+\log(6-x)}dx$

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

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