JEE Main 2021 (Online) 25th July Evening ShiftQuestion 1
The value of the integral $\int_{-1}^{1}\log\left(x+\sqrt{x^{2}+1}\right)dx$ is :
Definite Integration
Solution
Concept. $\log\big(x+\sqrt{x^2+1}\big)$ is odd, so its integral over $[-1,1]$ is $0$. Formulas used. (1) $\big(x+\sqrt{x^2+1}\big)\big(-x+\sqrt{x^2+1}\big)=1$Step 1 - Odd. By (1), $\log\big(-x+\sqrt{x^2+1}\big)=\log\frac1{x+\sqrt{x^2+1}}=-\log\big(x+\sqrt{x^2+1}\big)$. Step 2 - Value. $\int_{-1}^1(\text{odd})=0$. Answer: (A)