I was given this integral:
$$\int\frac{\ln {(1-t)}}{3t}\,\mathrm dt$$
I am having trouble with writing this as a power series, and I'm not sure where to start. I know I need the Maclaurin expansion of $\ln {(1-t)}$, but I'm not sure how to do this. Thanks!
Since$$\frac1{1-t}=1+t+t^2+t^3+\cdots,$$you have$$\log(1-t)=-\left(t+\frac{t^2}2+\frac{t^3}3+\cdots\right)$$and therefore\begin{align}\int\frac{\log(1-t)}{3t}\,\mathrm dt&=-\frac13\int1+\frac t2+\frac{t^2}3+\cdots\,\mathrm dt\\&=-\frac13\left(t+\frac{t^2}{2^2}+\frac{t^3}{3^2}+\cdots\right).\end{align}