Explanation

(a)
Let $f(x) = \log\left( x + \sqrt{1 + x^{2}} \right)$ and Replacing $x\ by - x,$ we get
$f( - x) = \log\left( \sqrt{1 + x^{2}} - x \right)$
$= \log\left( \sqrt{1 + x^{2}} - x \right)\frac{\left( \sqrt{1 + x^{2}} + x \right)}{\left( \sqrt{1 + x^{2}} + x \right)}$
$= \log\frac{\left\lbrack \left( 1 + x^{2} \right) - x^{2} \right\rbrack}{\left( \sqrt{1 + x^{2}} + x \right)}$
$= \log{1 - \log\left( \sqrt{1 + x^{2}} + x \right)}$
$= - \log\left( \sqrt{1 + x^{2}} + x \right)$
$\Rightarrow f( - x) = - f(x)$
Hence, $f(x)$ is an odd function.
$\therefore\ \int_{- 1}^{1}{\log{\left( x + \sqrt{1 + x^{2}} \right)dx = 0}}$