So you have (1/x)dx = -(1/(vlnv-v) )dv= -1/(v(lnv-1)
Now the integral of -1/(v(lnv-1) is -ln(lnv - 1)
Then proceed accordingly, hope I helped
Yes, that is the idea. However, just in case an overview helps...
... where (key in spoiler) ...
Spoiler:
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Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!