Question

The value of the integral $\underset{-1}{\overset{1}{\int }}\log (x+\sqrt{x^2+1})dx$ is

• A. $0$
• B. $-1$
• C. $2$
• D. $1$

Answer 1

The correct option is A $0$

$\underset{-1}{\overset{1}{\int }}\log (x+\sqrt{x^2+1})dx=\underset{-1}{\overset{1}{\int }}f\left(x\right)dx$
$\therefore f\left(x\right)=\log (x+\sqrt{x^2+1})$
$\Rightarrow f(-x)=\log (-x+\sqrt{x^2+1})$
$\Rightarrow f(-x)=\log \left(\right(-x+\sqrt{x^2+1})\times \frac{(x+\sqrt{x^2+1})}{(x+\sqrt{x^2+1})})$
$\Rightarrow f(-x)=\log \left(\frac{-x^2+x^2+1}{(x+\sqrt{x^2+1})}\right)$
$\Rightarrow f(-x)=-\log \left(x+\sqrt{x^2+1}\right)$
$\Rightarrow f\left(x\right)$ is odd function.
$\therefore \underset{-a}{\overset{a}{\int }}f\left(x\right)=0$
Hence, $\underset{-1}{\overset{1}{\int }}\log (x+\sqrt{x^2+1})dx=0$