Mathematics

# The integral $\displaystyle \int_2^4\frac {log x^2}{log x^2+log (36-12x+x^2)}dx$ is equal to

$1$

##### SOLUTION
$I = \displaystyle \int_2^4 { \dfrac{2 log x } { 2 log x + 2 log (6-x) } } dx$

$I = \displaystyle \int_2^4{ \dfrac{log x}{ log x + log (6-x)} }dx$

Using the property $\displaystyle \int_{a}^b f(x) \ dx= \int _{a}^{b} f(a+b-x) \ dx$
we get,

$I = \displaystyle \int_2^4{\dfrac{log (6-x)}{log x + log (6-x)} }dx$

Adding the two integrals we get,

$2 I = \displaystyle \int_2^4 {1}{dx}$

Hence, $2I = 2$

Hence, $I =1$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

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