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Here's one way to guess it: There is (a little known?) continuation process for linear differential equations, where at each step f^(k)/f^(n+1) is introduced: https://twitter.com/johncarlosbaez/status/1299442096621760512
This is possible, because at each stage, the derivative satisfies another linear equation, whose coefficients Q_k and P_k fulfill a "simple" recurrence.
The trick is to find such equation for the integral in question, for example:
This gives very simple coefficients of Q_k = -z/(k+1), P_k =1/(2k+2), and the fraction (after multiplying by denominators of Q):
Substituting for g(z) the integral reappears, and I hope you can manage the final tidying up of signs and fractions ;)
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