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829

Part of 2012 Today's Calculation Of Integral

Problems(1)

Today's calculation of Integral 829

Source: 2012 Kyoto Institute of Technology entrance exam

7/1/2012
Let aa be a positive constant. Find the value of ln⁡a\ln a such that ∫1eln⁡(ax) dx∫1ex dx=∫1eln⁡(ax)x dx.\frac{\int_1^e \ln (ax)\ dx}{\int_1^e x\ dx}=\int_1^e \frac{\ln (ax)}{x}\ dx.
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