Question

Solve:
$x\sqrt{1+x^2}+\log \left(x+\sqrt{x^2+1}\right).$

Answer 1

$x\sqrt{1+x^2}+log(x+\sqrt{1+x^2})$

$x+\sqrt{1+x^2}>0$

$\Rightarrow x\ge 0$___(1)

$x\sqrt{1+x^2}+log\left(\frac{(x+\sqrt{1+x^2})(x+\sqrt{1+x^2})}{(x-\sqrt{1+x^2})}\right)$

$\Rightarrow x\sqrt{1+x^2}+log\left(\frac{x^2-1-x^2}{x\sqrt{1+x^2}}\right)$

$\Rightarrow x\sqrt{1+x^2}+log\left(\frac{1}{\sqrt{1+x^2}-x}\right)$

$\frac{1}{\sqrt{1+x^2}-x}>0$

$\Rightarrow \sqrt{1+x^2}-x>0$

$\sqrt{1+x^2}>x$

$\Rightarrow 1+x^2>x^2$ $\to$ possible for all $xϵR$___(2)

$\therefore$ From (1) and (2)

$xϵ[0,\infty ]$