Yeah. But it's got a Taylor's series:
<table align="center" width="100%" cellpadding="0" cellspacing="0" style="padding-left: 50px" border="0"><tr style=""><td align="right" width="1"><img src="http://mathworld.wolfram.com/images/equations/TaylorSeries/inline22.gif" width="26" height="18" alt="lnx" /></td><td align="center" width="14"><img src="http://mathworld.wolfram.com/images/equations/TaylorSeries/inline23.gif" width="14" height="18" alt="=" /></td><td align="left"><img src="http://mathworld.wolfram.com/images/equations/TaylorSeries/inline24.gif" width="238" height="37" alt="lna+(x-a)/a-((x-a)^2)/(2a^2)+((x-a)^3)/(3a^3)-..." /></td></tr></table>
As for the integral of lnx, it's xlnx-x+c using Mountain.Dew's method.
You could do it without Taylor's series
∫lnxdx=∫lnx(dx/dx)dx
=xlnx-∫x(dlnx/dx)dx
=xlnx-∫x(1/x)dx
=xlnx-∫dx
=xlnx-x+c
Note that Taylor's series wasn't first discovered by Taylor. It was attributed to Brook Taylor who published it in 1715 in
Methodus in crementorum directa et inversa. But James Gregory wrote it on the back of a letter from a bookseller on January 30, 1671 (preserved in the library of the University of St. Andrews).
Taylor:
Gregory:
Maclaurin: