If a > 0 then  int_(2a)^(a^2)1/xdx=? 

Explanation:

Another Explanation (5): ```html Let \(I = \int_{2a}^{a^2} \frac{1}{x} dx\). Given \(a > 0\). We know that \(\int \frac{1}{x} dx = \ln |x| + C\). So, \(I = \left[ \ln |x| \right]_{2a}^{a^2} = \ln |a^2| - \ln |2a|\) Since \(a > 0\), we have \(|a^2| = a^2\) and \(|2a| = 2a\). Therefore, \(I = \ln (a^2) - \ln (2a)\). Using the property of logarithms, \(\ln(x) - \ln(y) = \ln(\frac{x}{y})\), we get: \(I = \ln \left( \frac{a^2}{2a} \right)\) \(I = \ln \left( \frac{a}{2} \right)\) \(I = \log_e \left( \frac{a}{2} \right)\) So, \(\int_{2a}^{a^2} \frac{1}{x} dx = \log_e \frac{a}{2}\). 😊 ```