What is the integral of #[ln(lnx)]/[x] dx#?
1 Answer
Apr 26, 2015
I got:
#lnx*ln(lnx) - lnx + C#
Now we can do some unusual substitution...
First, recognize that
#f(u) = ln(u)du#
where#u = lnx# and#du = 1/xdx# .
Further considerations lead to using integration by parts on
#s = lnu#
#ds = 1/udu#
#dt = du#
#t = u#
#intln(u)du#
#= st - int tds#
#= u*lnu - int u*1/udu#
#= u*lnu - u#
#= color(blue)(lnx*ln(lnx) - lnx + C)#