P(x)= 1/x，q(x)= e x/x∫p(x)dx = lnxy =( 1/x)[∫(e x/x)* x dx+c]=(。 Then the molecule is also →0, that is, e 0+c = 0c =- 1, so y = (e x- 1)/x, that is, F (X) = (EX-1)/XF (X) = F' (X) = + 1/x？ Because1/x =-1[1-(x+1)] =-∑ (n = 0, ∞) (x+1) n1/x. =(- 1/x)'=∑(n=0，∞)n(x+ 1)^(n- 1)e^x=∑(n=0,∞)x^n/n！ So f (x) = ∑ (n = 0, ∞) x (n- 1)/n! - ∑(n=0，∞) x^(n-2)/n！ +∑(n=0，∞)n(x+ 1)^(n- 1)n/(n+ 1)！ =(n+ 1- 1)(n+ 1)！ =(n+ 1)/(n+ 1)！ - 1/(n+ 1)！ = 1/n！ - 1/(n+ 1)！ Then the final conclusion in the process of summation is ∑(n= 1, ∞)n/(n+ 1)! = 1/ 1 - 1/2! + 1/2! - 1/3! +……= 1- 1/(n+ 1)！ = 1