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A u sub of u equal to ln(ln) would have been more effective here wouldn't it?
u = ln(lnx)du = 1/x * 1/lnx dxdx = xlnx duwhen x = e, u = ln(1) = 0when x = e^2, u = ln(ln(e^2)) = ln(2)∫e^{e^2} (ln(lnx))/(xlnx) dx= ∫0^ln(2) u/(xlnx) * xlnx du= ∫0^ln(2) u du= 1/2[u^2]0^ln2= 1/2(ln2)^2
A u sub of u equal to ln(ln) would have been more effective here wouldn't it?
u = ln(lnx)
du = 1/x * 1/lnx dx
dx = xlnx du
when x = e, u = ln(1) = 0
when x = e^2, u = ln(ln(e^2)) = ln(2)
∫e^{e^2} (ln(lnx))/(xlnx) dx
= ∫0^ln(2) u/(xlnx) * xlnx du
= ∫0^ln(2) u du
= 1/2[u^2]0^ln2
= 1/2(ln2)^2