Question

The value of integral $\underset{3}{\overset{5}{\int }}\frac{\ln x^2}{\ln x^2+\ln (64-16x+x^2)}dx,$ is

• A. $0$
• B. $1$
• C. $2$
• D. $2\ln 5-\ln 3$

Answer 1

The correct option is B $1$

Let $I=\underset{3}{\overset{5}{\int }}\frac{\ln x^2}{\ln x^2+\ln (64-16x+x^2)}dx=\underset{3}{\overset{5}{\int }}\frac{\ln x^2}{\ln x^2+\ln (x-8{)}^2}dx\cdots \left(i\right)$
Using the property:
$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$\Rightarrow I=\underset{3}{\overset{5}{\int }}\frac{\ln (x-8{)}^2}{\ln x^2+\ln (x-8{)}^2}dx\cdots \left(ii\right)$
adding $\left(i\right)$ and $\left(ii\right):$
$\Rightarrow 2I=\underset{3}{\overset{5}{\int }}1dx$
$\Rightarrow I=\frac{1}{2}[x{]}_3^5=1$

Alternate solution : As $f\left(x\right)+f(a+b-x)=1$
$\Rightarrow I=\frac{b-a}{2}=\frac{5-3}{2}=1$