Question

The integral $\underset{2}{\overset{4}{\int }}\frac{\log x^2}{\log x^2+\log \left(36-12x+x^2\right)}dx$ is equal to

• A. 2
• B. 4
• C. 1
• D. 6

Answer 1

Answer: (C)

PLAN Apply the property
$\underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$ and then add.
Let $=\underset{2}{\overset{4}{\int }}\frac{\log x^2}{\log x^2+\log \left(36-12x+x^2\right)}dx$
$=\underset{2}{\overset{4}{\int }}\frac{2\log x}{2\log x+\log (6-x{)}^2}dx=\underset{2}{\overset{4}{\int }}\frac{2\log xdx}{2[\log x+\log (6-x)]}$
$\Rightarrow I=\underset{2}{\overset{4}{\int }}\frac{\log xdx}{[\log x+\log (6-x)]}....$(i)
$\Rightarrow I=\underset{2}{\overset{4}{\int }}\frac{\log (6-x)}{\log (6-x)+\log x}dx....$(ii)
$\left[\because \underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx\right]$
On adding Eqs. (i) and (ii), we get
$2I=\underset{2}{\overset{4}{\int }}\frac{\log x+\log (6-x)}{\log x+\log (6-x)}dx$
$\Rightarrow 2I=\underset{2}{\overset{4}{\int }}dx=[x{]}_2^4\Rightarrow 2I=2$
$\Rightarrow 2I=2\Rightarrow I=1$