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.

  • 1
    $\begingroup$ Map[Integrate[#, x] &, x^2 + (Log[x] Log[1 - x])^2] $\endgroup$
    – Bill
    Apr 25, 2017 at 18:46

1 Answer 1


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

  • $\begingroup$ 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, 2017 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, 2017 at 19:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.