First we find the series representation of the basic, 1/(1-x). If you already know the answer, it is 1+x+x^2+x^3+x^4...., easy to remember. If not, we can use the binomial expansion: 1/(1-x) = (1-x)^(-1) = 1+((-1)/1!)(-x)+(-1)(-2)/2!(-x)^2+(-1)(-2)(-3)/3!(-x)^3+... which gives 1+x+x^2+x^3+x^4+...
Then 1/(1-x)^2 is just (1/(1-x))^2, or (1+x+x^2+x^3+x^4+...)^2= x+x^2+x^3+x^4+x^5+x^6+... x^2+x^3+x^4+x^5+x^6+... +x^3+x^4+x^5+x^6+... +x^4+x^5+x^6+... +x^5+x^6+... .... ....) =1+2x+3x^2+4x^3+5x^4+6x^5+....
Similarly, (1+x)/(1-x)^2 can be considered as =1/(1+x)^2+x(1+x)^2 =1+2x+3x^2+4x^3+5x^4+6x^5+.... +x+2x^2+3x^3+4x^4+5x^5+... =1+3x+5x^2+7x^3+9x^4+11x^5+...
Finally, x(1+x)/(1-x)^2 can be considered as =x*(1+x)/(1-x)^2 =x*(1+3x+5x^2+7x^3+9x^4+11x^5+...) =x+3x^2+5x^3+7x^4+9x^5+11x^6+....+(2i-1)x^i+...
Therefore x(1+x)/(1-x)^2 = x+3x^2+5x^3+7x^4+9x^5+11x^6+...+(2i-1)x^i+... ad infinitum
