The set of all real numbers for which 
\[x+\sqrt{x^2+1}-\frac{1}{x+\sqrt{x^2+1}}\]
is a rational number is the set of all
(A) integers $x$  (B) rational $x$  (C) real $x$
(D) $x$ for which $\sqrt{x^2+1}$ is rational
(E) $x$ for which $x+\sqrt{x^2+1}$ is rational
Rationalizing the denominator of $\frac{1}{x+\sqrt{x^2+1}}$, it simplifies: $\frac{1}{x+\sqrt{x^2+1}}$ = $\frac{x-\sqrt{x^2+1}}{x^2-(x^2+1)}$ = $\frac{x-\sqrt{x^2+1}}{-1}$ = $-(x-\sqrt{x^2+1})$. Substituting this into the original equation, we get $x + \sqrt{x^2+1} - (-(x-\sqrt{x^2+1})) = x+\sqrt{x^2+1} + x - \sqrt{x^2+1} = 2x$. $2x$ is only rational if $x$ is rational $\mathrm{(B)}$