Mathematics

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

$1$

##### SOLUTION
$I=\int _{ 2 }^{ 4 }{ \cfrac { \log { { x }^{ 2 } } }{ \log { { x }^{ 2 } } \log { (36-12x+{ x }^{ 2 }) } } dx } =\int _{ 2 }^{ 4 }{ \cfrac { \log { { x }^{ 2 } } }{ \log { { x }^{ 2 } } { (6-x) }^{ 2 } } dx } \quad -(i)\\ \int _{ a }^{ b }{ f(x) } dx=\int _{ 2 }^{ 4 }{ f(a+b-x)dx } \\ \therefore I=\int _{ 2 }^{ 4 }{ \cfrac { \log { { (4+2-x) }^{ 2 } } }{ \log { { (6-x) }^{ 2 } } +\log { { (6-6+x) }^{ 2 } } } dx } \\ I=\int _{ 2 }^{ 4 }{ \cfrac { \log { { (6-x) }^{ 2 } } }{ \log { { (6-x) }^{ 2 } } +\log { { (6-6+x) }^{ 2 } } } dx } \\ I=\int _{ 2 }^{ 4 }{ \cfrac { \log { { (6-x) }^{ 2 } } }{ \log { { (6-x) }^{ 2 } } +\log { { x }^{ 2 } } } dx } \quad -(ii)$
Adding $(i)$ & $(ii)$
$2I=\int _{ 2 }^{ 4 }{ \cfrac { \log { { (6-x) }^{ 2 } } +\log { { x }^{ 2 } } }{ \log { { (6-x) }^{ 2 } } +\log { { x }^{ 2 } } } dx } =\int _{ 2 }^{ 4 }{ dx } ={ x }_{ 2 }^{ 4 }=(4-2)=2\\ 2I=2\\ \Rightarrow I=1$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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