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Integrate seems to return a whole expression unresolved whenever just a single term is unintegrable. How do I get it to return the integral of all terms that are resolvable with the only the unintegrable terms unresolved?

For example, I think Integrate[x^2 + (Log[x]Log[1-x])^2,x] should return x^3/3 + Integrate[(Log[x]Log[1-x])^2,x]; instead it returns the whole expression unintegrated. Even invoking Simplify or FullSimplify doesn't help.

I'm Integrating an expression that, when expanded, has only one unintegrable term, but I would like to have the results for the other terms.

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    $\begingroup$ Map[Integrate[#, x] &, x^2 + (Log[x] Log[1 - x])^2] $\endgroup$ – Bill Apr 25 '17 at 18:46
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Map the Integrate onto the expression. "Map[f,expr] or f/@expr applies f to each element on the first level in expr."

Integrate[#, x] & /@ (x^2 + (Log[x] Log[1 - x])^2)

enter image description here

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  • $\begingroup$ I appreciate the pointer to the Map function. I'm just getting back to using Mathematica after about a 15 year hiatus, so have to lot to catch up on. $\endgroup$ – wjv3 Apr 25 '17 at 19:07
  • $\begingroup$ But the question remains: Why aren't the linear properties of the Integral operator incorporated into the function itself? Is their some subtle advantage or some subtle problem avoided by not using this property? $\endgroup$ – wjv3 Apr 25 '17 at 19:10

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