Verified answer

Assuming expansion about x = 1:

Start with the geometric series:

1/(1 - t) = Σ(n = 0 to ∞) t^n.

Let t = x - 1:

-1/x = Σ(n = 0 to ∞) (x - 1)^n.

==> 1/x = -Σ(n = 0 to ∞) (x - 1)^n.

Integrate both sides from 1 to x:

ln x = -Σ(n = 0 to ∞) (x - 1)^(n+1)/(n+1).

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Divide both sides by x - 1:

ln(x)/(x - 1) = -Σ(n = 0 to ∞) (x - 1)^n/(n+1).

Integrate both sides:

∫ ln x dx/(x - 1) = C - Σ(n = 0 to ∞) (x - 1)^(n+1)/(n+1)^2.

I hope this helps!