F feyomi Jan 2010 50 0 May 26, 2010 #1 I'm struggling to find a suitable substitution. If I know one then I should be able to work it out from there. Thanks for any help.

I'm struggling to find a suitable substitution. If I know one then I should be able to work it out from there. Thanks for any help.

General Jan 2010 564 242 Kuwait May 26, 2010 #2 put \(\displaystyle u=ln(x) \implies du=\frac{dx}{x}\) So the integral will be : \(\displaystyle \int \frac{du}{u^2}=\frac{-1}{u}+C=\frac{-1}{ln(x)} + C\) Reactions: HallsofIvy and feyomi

put \(\displaystyle u=ln(x) \implies du=\frac{dx}{x}\) So the integral will be : \(\displaystyle \int \frac{du}{u^2}=\frac{-1}{u}+C=\frac{-1}{ln(x)} + C\)