How do you find #int x*secxdx#?

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Answer 1 · 2016-11-29T13:31:10

That cannot be evaluated by introductory level functions.

The integral involves the Polylogarithm function #Li_2(x) = sum_(k=1)^oo x^k/k^2#.

WolframAlpha gives

#int x secx dx = i(Li_2(-ie^(ix))-Li_2(ie^(ix)))+x(ln(1-ie^(ix))-ln(1+ie^(ix))) +C#

You can read more about the polylogarithm here: http://mathworld.wolfram.com/Polylogarithm.html