Power Function . . .?

Question

Evaluate the indefinite integral as a power series:

Int [ ln(1-t) / t ] dt

Answer 1

Ln(1 - x) =
∑ [(-1)^n(-x)^(n + 1)]/(n + 1) =
∑ [(-1)^n(-1)^(n + 1)(x)^(n + 1)]/(n + 1) =
∑ [(-1)(x)^(n + 1)]/(n + 1) =
∫[(Ln(1 - t))/t]dt =
∫∑ [- t^n]/(n + 1) dt =
- (C + t/1 + t^2/4 + t^3/9 + t^4/16 + . . .)

Answer 2

You first need to find the power series for ln(1 - t). This you can find in a table of standard power series and it is
ln(1 - t) = -t - (t^2)/2 - (t^3)/3 - (t^4)/4 . . . (-1 < t < 1)
You then divide each term by t and integrate term by term. There may be some problem with the convergence of the integrated power series.