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find the integral of ln(lnx)/xlnx
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Let u = ln(lnx). derivative of lnu is u'/u, so ln(lnu) is the derivative of ln(u) divided by lnu, so that means du = \[\frac{ \frac{ 1 }{ x } }{ lnx }=\frac{ 1 }{ xlnx }dx\] Meaning dx = (du)(xlnx) \[\frac{ u }{ xlnx }*du(xlnx)\]Leaving us finally with simply \[\int\limits_{}^{}udu=\frac{ u^{2} }{ 2 }\]And since u = ln(lnx), the final answer is: \[\frac{ (\ln(lnx))^{2} }{ 2 }+C\]
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