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We have : $$\int \frac{1}{x} \sum_{m=1}^\infty \frac{(x y)^m}{1-y^m} dx = \sum_{m=1}^\infty \frac{y^m}{1-y^m} \int x^{m-1}dx= \sum_{m=1}^\infty \frac{y^m}{1-y^m} \frac{x^m}{m} = \sum_{m=1}^\infty \frac{(x y)^m}{(1-y^m)m}$$

From Sotiris' answer to a previous question here, based off: Jjacqueline's answer to a similar question here. I think I could follow and maybe improve this if I understood a bit more about how we swap out the integral and the summation signs in the line I mentioned above. I get up to: $$\sum_{m=1}^\infty \int{\frac{y^m}{1-y^m} x^{m-1}}dx$$ How does $$\int \frac{1}{x} \sum_{m=1}^\infty \frac{(x y)^m}{1-y^m} dx$$ come from that? I ask because I don't understand why multiple integrations would be needed in the case of the $\frac{1}{m!}$ hypothetical I asked Sotiris about in the first link of this question, knowing the partial derivative of $\frac{x^m}{m!}$ with respect to $m$ is this.

user3108815
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I somehow forgot that the integral of polynomial summands is just the integral of the sum of the terms.

i.e: $$if\ f(i) = a_i,$$ and $$g(i) = a_ix^i= f(i)x^i,$$ then $$\sum_{i=1}^n\bigr(\int a_ix^i\ dx\bigl)\ =\int a_1x\ dx+\int a_2x^2\ dx + \int a_3x^3\ dx+...+\int a_nx^n\ dx$$ $$ =\int a_1x\ + a_2x^2\ + a_3x^3\ +...+a_nx^n\ dx$$ $$= \int (\sum_{i=1}^{n}f(i)x^i)\ dx$$ $$= \int \bigl(\sum_{i=1}^{n}g(i)\bigr)\ dx$$

user3108815
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