Question

$\underset{6}{\overset{16}{\int }}\frac{{\log }_e{x}^2}{{\log }_e{x}^2+{\log }_e(x^2-44x+484)}dx$ is equal to

• A. $6$
• B. $10$
• C. $5$
• D. $8$

Answer 1

The correct option is C $5$

Let $I=\underset{6}{\overset{16}{\int }}\frac{{\log }_e{x}^2}{{\log }_e{x}^2+{\log }_e(x^2-44x+484)}dx$
$\Rightarrow I=\underset{6}{\overset{16}{\int }}\frac{2\ln x}{2\ln x+2\ln (22-x)}dx$
$\Rightarrow I=\underset{6}{\overset{16}{\int }}\frac{\ln x}{\ln x+\ln (22-x)}dx\cdots \left(i\right)$
Using the property, $I=\underset{a}{\overset{b}{\int }}\frac{f\left(x\right)}{f\left(x\right)+f(a+b-x)}dx=\underset{a}{\overset{b}{\int }}\frac{f(a+b-x)}{f(a+b-x)+f\left(x\right)}dx$
$\therefore I=\underset{6}{\overset{16}{\int }}\frac{\ln (22-x)}{\ln (22-x)+\ln x}dx\cdots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right),$ we get
$2I=\underset{6}{\overset{16}{\int }}dx=10$
$\therefore I=5$