Question

${\int }_{-1}^1\log (x+\sqrt{x^2+1})dx$ is equal to

• A. $0$
• B. $\log 2$
• C. $\log \frac{1}{2}$
• D. None of these

Answer 1

The correct option is A $0$

Let $f\left(x\right)=\log (x+\sqrt{1+x^2})$
$\therefore f(-x)=\log \{-x+\sqrt{1+\left({-x}^2\right)}\}$
$=\log (\sqrt{1+x^2}-x)\times \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}+x}$
$=\log \left(\frac{1+x^2-x^2}{\sqrt{1+x^2}+x}\right)$
$=\log {(\sqrt{1+x^2}+x)}^{-1}$
$=-\log (\sqrt{1+x^2}+x)$
$=-f\left(x\right)$
Thus, $f(-x)=-f\left(x\right)$
$\therefore f\left(x\right)$ is an odd function.
So, ${\int }_{-1}^1\log (x+\sqrt{1+x^2})dx=0$