- #1
JustaNickname
- 4
- 0
∫dx/(x^2*lnx)
What I`ve seen on the web but I don`t 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!
What I`ve seen on the web but I don`t 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!