JEE Main 2021 (Online) 4th August Evening ShiftQuestion 1
The value of the integral $\displaystyle\int_{1}^{6}\dfrac{\log_{e}x\,dx}{\log_{e}x^{2}+\log_{e}\left(x^{2}-14x+49\right)}$ is
Definite Integration
Solution
Concept. The denominator is $2\ln x+2\ln(7-x)$; the substitution $x\to7-x$ swaps the two logs. Formulas used. (1) $\ln(x^2-14x+49)=\ln(7-x)^2=2\ln(7-x)$ on $[1,6]$(2) $\int_a^bg(x)\,dx=\int_a^bg(a+b-x)\,dx$Step 1 - Simplify. By (1), the integral is $\frac12J$ with $$J=\int_1^6\frac{\ln x}{\ln x+\ln(7-x)}dx.$$ Step 2 - Use (2). $a+b=7$: $J=\int_1^6\frac{\ln(7-x)}{\ln(7-x)+\ln x}dx$. Adding, $2J=\int_1^61\,dx=5$. Step 3 - Value. $J=\frac52$, so the integral is $\frac54$. Answer: (A)