Interesting Summation!

Nice problem and here is my solution

For $|x| < 1$ , $f(x)$ can be written as $\displaystyle \ln(1-x)+\sum_{k=1}^{n}\frac{x^k}{k}=-\sum_{k=1}^{\infty} \frac{x^{k+n}}{k+n}=-\sum_{k=1}^{\infty}\int_0^x u^{k+n-1}du=-\int_0^x\frac{u^n}{1-u}du$ and hence we have $-\sum_{n\geq 1}(-1)^{n-1} n f(x)=-\sum_{n\geq 1}(-1)^{n-1} n\int_0^x \frac{u^n}{1-u} du\\ =-\int_0^x\frac{du}{1-u}\left(\sum_{n\geq 1}\ (-1)^{n-1} n u^n \right)=-\int_0^x\frac{du}{1-u} \left( u\cdot \frac{d}{du}\frac{1}{(1+u)}\right)\\=\int_0^x\frac{-u}{(1-u)(1+u)^2} du =I(x)$ and hence we are left to evaluate $I(x)=$ . $\int_0^x\left(\frac{1}{(1+u)^2}-\frac{1}{(1-u)(1+u)^2}\right)du$ set $x=\frac{1}{2}$ and it's easy to integrate the above two integral ie $\int_0^{\frac{1}{2}}\frac{du}{(1+u)^2}=-\frac{2}{3}+1=\frac{1}{3}$ and on the similar way $\int_0^{\frac{1}{2}}\frac{du}{(1-u)(1+u)^2}=\left[\frac{1}{4}\ln\left(\frac{1+u}{1-u}\right)+\frac{1}{2(1+u)}\right]_0^{\frac{1}{2}}=\frac{1}{4}\ln3+\frac{1}{6}$ and hence we have $I(2^{-1})=\frac{1}{3}-\left(\frac{1}{6}+\frac{1}{4}\ln 3 \right)=\frac{1}{6}-\frac{1}{4}\ln 3$ making our required answer $abcd=72$