The expansion of 1/√ (1+x)+√ (1−x) in ascending powers of x, up to and including the term in x2

Via conjugates,
1 / [√(1 + x) + √(1 - x)]
= [√(1 + x) - √(1 - x)] / {[√(1 + x) + √(1 - x)] [√(1 + x) - √(1 - x)]}
= [√(1 + x) - √(1 - x)] / [(1 + x) - (1 - x)]
= (1/(2x)) * [√(1 + x) - √(1 - x)].

Now, use the binomial series with exponent 1/2:
√(1 + x) = 1 + (1/2)x + (1/2)(1/2 - 1) x^2/2! + (1/2)(1/2 - 1)(1/2 - 2) x^3/3! + ...
.............= 1 + (1/2)x - (1/8) x^2 + (1/16) x^3 + ...

Replacing x with -x:
√(1 + x) = 1 - (1/2)x - (1/8) x^2 - (1/16) x^3 + ...

Hence, (1/(2x)) * [√(1 + x) - √(1 - x)]
= (1/(2x)) {[1 + (1/2)x - (1/8) x^2 + (1/16) x^3 + ...] - [1 - (1/2)x - (1/8) x^2 - (1/16) x^3 + ...]}
= (1/(2x)) [x + (1/8) x^3 + ...]
= 1/2 + (1/16) x^2 + ...

I hope this helps!