Question

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

• A. $4$
• B. $1$
• C. $6$
• D. $2$

Answer 1

Answer: (B)

$1$

$I={\int }_2^4\frac{\log x^2}{\log x^2+\log (36-12x+x^2)}dx....\left(1\right)$

$I={\int }_2^4\frac{\log {(4+2-x)}^2}{\log (4+2-x)+\log {[6-(4+2-x\left)\right]}^2}dx$

$={\int }_2^4\frac{\log {(6-x)}^2}{\log {(6-x)}^2+\log {\left(x\right)}^2}dx[\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(b+a-x)dx]....\left(2\right)$

Adding (1) and (2)

$2I={\int }_2^4\frac{\log {\left(x\right)}^2+\log {(6-x)}^2}{\log {\left(x\right)}^2+\log {(6-x)}^2}dx={\left[x\right]}_2^4=2$