I am attempting to show that $$ \sum_{k \ge 1}^\infty {k^2 x^k \over (1+x^k)^2} \sim (1-x)^{-3} {\pi^2 \over 6} $$ as $x$ approaches 1 from below. The sum can be approximated by the integral $$ \int_0^\infty {k^2 x^k \over (1+x^k)^2} \: dk $$ and making the change of variable $u = x^k$ this is equal to $$ {1 \over \log^3 x} \int_1^0 {\log^2 u \over (1+u)^2} \: du. $$ Maple tells me that this integral is equal to $\pi^2/6$. Moreover if I try to find the same integral but with arbitrary rational numbers as bounds, it is successful: for example $$ \int_{1/2}^2 {\log^2 u \over (1+u)^2} \: du = -4 \log 2 \log 3 + {7 \over 3} \log^2 2 - 2 dilog(3) + 2 dilog(3/2). $$ Specifically, I'd like to know:

How can I find the integral in the third displayed equation (or,

*really*what I'm interested in, the sum in the first displayed equation)?How can I find the

*indefinite*integral $\int \log^2 u / (1+u)^2 \: du$?Various integrals of this type (roughly speaking, logarithms divided by polynomials) are recurring in my work. Presumably they can be computed with the ordinary techniques of calculus and the correct identities involving polylogarithms. What are these identities, or where can I find them?