Answer
$\sqrt{x^{2}+1}-x=\dfrac{1}{\sqrt{x^{2}+1}+x}$
Work Step by Step
$\sqrt{x^{2}+1}-x$
Multiply this expression by $\dfrac{\sqrt{x^{2}+1}+x}{\sqrt{x^{2}+1}+x}$ and simplify:
$\sqrt{x^{2}+1}-x=(\sqrt{x^{2}+1}-x)\cdot\dfrac{\sqrt{x^{2}+1}+x}{\sqrt{x^{2}+1}+x}=...$
$...=\dfrac{(\sqrt{x^{2}+1})^{2}-x^{2}}{\sqrt{x^{2}+1}+x}=\dfrac{x^{2}+1-x^{2}}{\sqrt{x^{2}+1}+x}=\dfrac{1}{\sqrt{x^{2}+1}+x}$
