Negative binomial expansion problem

Question

So here's the problem:-

Show that if x is small, the expression $$\sqrt{(1+x)(1-x)^{-1}}\approx 1+x+0.5x^2$$

Answer 1

HInt:

$$\sqrt{(1+x)(1-x)^{-1}}=(1+x)^{1/2}(1-x)^{-1/2}$$

Or $$\sqrt{(1+x)(1-x)^{-1}}=(1+x)(1-x^2)^{-1/2}$$

Now use Binomial series, $$(1+x)^n=1+nx+\dfrac{n(n-1)}2x^2+\cdots$$ for $|x|<1$

Answer 2

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

$$=\sqrt{1+2x+2x^2+\ldots}=1+\frac{2x+2x^2+\ldots}2-\frac{(2x+2x^2+\ldots)^2}8+\ldots=$$

$$=1+x+\left(x^2-\frac{4x^2}8\right)+\ldots=1+x+\frac12x^2+\ldots$$