Answer
$\frac{1}{2}x+\Sigma_{n=1}^{\infty}(-1)^{n}\frac{1.3.5....(2n-1)x^{(2n+1)}}{2^{(3n+1)}n!}$
$R=2$
Work Step by Step
$\frac {x}{\sqrt{(4+x^{2})}}=\frac{x}{2}\frac{1}{\sqrt {1+\frac{x^{2}}{4}}}$
$=\frac{x}{2}(1+\frac{x^{2}}{4})^{-1/2}$
$= \frac{x}{2}+(-\frac{1}{2})(\frac{x}{2})^{3}+\frac{(\frac{-1}{2})(\frac{-3}{2})}{2!}(\frac{x}{2})^{5}+\frac{(\frac{-1}{2})(\frac{-3}{2})(\frac{-5}{2})}{3!}(\frac{x}{2})^{7}+....$
$=\frac{1}{2}x-\frac{1}{16}x^{3}+\frac{3}{256}x^{5}-....$
$=\frac{1}{2}x+\Sigma_{n=1}^{\infty}(-1)^{n}\frac{1.3.5....(2n-1)x^{(2n+1)}}{2^{(3n+1)}n!}$
$R=2$
