# Integral of 1/(x^2*lnx) !

1. Feb 20, 2012

### JustaNickname

∫dx/(x^2*lnx)

What Ive seen on the web but I dont think is right:
u= lnx *** what we have here isn't lnx but (lnx)^-1... This is why I doubt that's the right solution
du = dx/x
dv = dx/x^2
v = -1/x

=-lnx/x + ∫dx/x^2
=-lnx/x - 1/x + C

Let me know if it is correct, thanks!

2. Feb 20, 2012

### Jorriss

I didn't check if it is correct myself, but you never really need someone to check if an integral (like this) is correct. Just differentiate it.

3. Feb 20, 2012

### Ray Vickson

Maple 14 gets a non-elementary function for this integral; that is, it cannot be done in terms of elementary functions of the type you have used.

RGV