Question

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

• A. $2$
• B. $4$
• C. $1$
• D. $6$

Answer 1

The correct option is C $1$

$I={\int }_2^4\frac{\log x^2}{\log x^2\log (36-12x+x^2)}dx={\int }_2^4\frac{\log x^2}{\log x^2{(6-x)}^2}dx-\left(i\right){\int }_a^{b}f\left(x\right)dx={\int }_2^{4}f(a+b-x)dx\therefore I={\int }_2^4\frac{\log {(4+2-x)}^2}{\log {(6-x)}^2+\log {(6-6+x)}^2}dxI={\int }_2^4\frac{\log {(6-x)}^2}{\log {(6-x)}^2+\log {(6-6+x)}^2}dxI={\int }_2^4\frac{\log {(6-x)}^2}{\log {(6-x)}^2+\log x^2}dx-\left(ii\right)$

Adding $\left(i\right)$ & $\left(ii\right)$

$2I={\int }_2^4\frac{\log {(6-x)}^2+\log x^2}{\log {(6-x)}^2+\log x^2}dx={\int }_2^{4}dx=x_2^4=(4-2)=22I=2\Rightarrow I=1$