Considering the general term of S(x)=sum_{i=2}^oo (x+1)^i /(i-1)^2S(x)=∞∑i=2(x+1)i(i−1)2 as
t_i = (x+1)^i/(i-1)^2ti=(x+1)i(i−1)2
we construct the following relations
d/(dx)(t_i/(x+1))=((x+1)^{i-2})/(i-1)ddx(tix+1)=(x+1)i−2i−1
d/(dx)((x+1)d/(dx)(t_i/(x+1)))=(x+1)^{i-2}ddx((x+1)ddx(tix+1))=(x+1)i−2
(x+1)^2d/(dx)((x+1)d/(dx)(t_i/(x+1)))=(x+1)^i(x+1)2ddx((x+1)ddx(tix+1))=(x+1)i
but
sum_{i=2}^{oo}(x+1)^i =- (1/x+2+x)∞∑i=2(x+1)i=−(1x+2+x) for -2 < x < 0−2<x<0
then
sum_i d/(dx)((x+1)d/(dx)(t_i/(x+1)))=-1/(x+1)^2 (1/x+2+x)∑iddx((x+1)ddx(tix+1))=−1(x+1)2(1x+2+x)
and
sum_id/(dx)(t_i/(x+1)) =-1/(x+1) int 1/(x+1)^2 (1/x+2+x) dx∑iddx(tix+1)=−1x+1∫1(x+1)2(1x+2+x)dx
proceeding this way we get
S(x)=sum_i t_i=-(x+1) (c_2 + c_1 Log_e(1 + x) + Log_e(-x)Log_e(1 + x) + "PolyLog"(2, -x))S(x)=∑iti=−(x+1)(c2+c1loge(1+x)+loge(−x)loge(1+x)+Polylog(2,−x))
Attached the comparisson between sum_{i=2}^oo (x+1)^i /(i-1)^2∞∑i=2(x+1)i(i−1)2 and S(x)S(x)
