Simplify 1/(sqrt(x^2+1))-(x^2)/((x^2+1)^(3/2))

Simplify $\frac{1}{\sqrt{{x}^{2}+1}}-\frac{{x}^{2}}{\left({x}^{2}+1{\right)}^{3/2}}$
I want to know why
$\frac{1}{\sqrt{{x}^{2}+1}}-\frac{{x}^{2}}{\left({x}^{2}+1{\right)}^{3/2}}$
can be simplified into
$\frac{1}{\left({x}^{2}+1{\right)}^{3/2}}$
I tried to simplify by rewriting radicals and fractions. I was hoping to see a clever trick (e.g. adding a clever zero, multiplying by a clever one? Quadratic completion?)
$\begin{array}{rl}\frac{1}{\sqrt{{x}^{2}+1}}-\frac{{x}^{2}}{\left({x}^{2}+1{\right)}^{3/2}}& =\\ & =\left({x}^{2}+1{\right)}^{-1/2}-{x}^{2}\ast \left({x}^{2}+1{\right)}^{-3/2}\\ & =\left({x}^{2}+1{\right)}^{-1/2}\ast \left(1-{x}^{2}\ast \left({x}^{2}+1{\right)}^{-1}\right)\\ & =...\end{array}$
To give a bit more context, I was calculating the derivative of $\frac{x}{\sqrt{{x}^{2}+1}}$ in order to use newtons method for approximating the roots.
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Manuel Leach
$\sqrt{x}$ is just a shorthand for ${x}^{1/2}$. Hence we can multiply the two halves of the first fraction in the first term by ${x}^{2}+1$:
$\frac{1}{\sqrt{{x}^{2}+1}}-\frac{{x}^{2}}{\left({x}^{2}+1{\right)}^{3/2}}=\frac{{x}^{2}+1}{\left({x}^{2}+1{\right)}^{3/2}}-\frac{{x}^{2}}{\left({x}^{2}+1{\right)}^{3/2}}$
and the target expression follows.

Gaintentavyw4
HINT: we have
$\frac{1}{\sqrt{{x}^{2}+1}}-\frac{{x}^{2}}{{\sqrt{{x}^{2}+1}}^{3}}=\frac{{x}^{2}+1-{x}^{2}}{{\sqrt{{x}^{2}+1}}^{3}}$