Let us first simplify
#sqrt(1-x)+sqrt((1+x)^2/(1-x))+(1+x)#
= #1/sqrt(1-x)[1-x+sqrt((1+x)^2)+(1+x)sqrt(1-x)]#
= #1/sqrt(1-x)[1-x+1+x+(1+x)sqrt(1-x)]#
= #(1-x)^(-1/2)[2+(1+x)(1-x)^(1/2)]#
= #2(1-x)^(-1/2)+(1+x)#
= #1+x+2(1-x)^(-1/2)#
We can now expand #(1-x)^(-1/2)# usng Binomial Theorem according to which
#(1-x)^n=1-nx+(n(n-1))/(2!)x^2+(n(n-1)(n-2))/(3!)x^3+(n(n-1)(n-2)(n-3))/(4!)x^3+...........#
Hence #(1-x)^(-1/2)=1-(-1)/2x+((-1/2)(-1/2-1))/(2!)x^2-((-1/2)(-1/2-1)(-1/2-2))/(3!)x^3+((-1/2)(-1/2-1)(-1/2-2)(-1/2-3))/(4!)x^3+...........#
= #1+1/2x+(1/2*3/2)/2x^2+(1/2*3/2*5/2)/6x^3+(1/2*3/2*5/2*7/2)/24x^4+........#
= #1+1/2x+3/8x^2+5/16x^3+35/128x^4+..........#
Hence #sqrt(1-x)+sqrt((1+x)^2/(1-x))+(1+x)=1+x+2(1-x)^(-1/2)#
= #1+x+2(1+1/2x+3/8x^2+5/16x^3+35/128x^4+.........)#
= #1+x+2+x+3/4x^2+5/8x^3+35/64x^4+.........#
= #3+2x+3/4x^2+5/8x^3+35/64x^4+.........#
and ignoring higher powers of #x#
#sqrt(1-x)+sqrt((1+x)^2/(1-x))+(1+x)~=3+2x# and not #(1+x/3)#
