f(x)=∑x^n/[n(n+1)]
求導(dǎo)：f'(x)=∑ x^(n-1)/(n+1)
F=x^2f'(x)= ∑ x^(n+1)/(n+1)
再求導(dǎo)：F'=∑ x^n=x/(1-x)=1/(1-x)-1
積分：F=-ln(1-x)-x
f'(x)=F/x^2=-ln(1-x)/x^2-1/x
再積分：f(x)=ln(1-x)/x+∫x/(1-x)dx-lnx=ln(1-x)/x-ln(1-x)-x-lnx=[ln(1-x)]/x-ln[x(1-x)]-x