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Evaluate y int e3a.log x + e3x.logadx

Given:\(\begin{array}{l}\int e^{3a.logx} + e^{3x.loga}dx\end{array} \) We know that \(\begin{array}{l}a^{mn} = (a^{m})^{n}\end{array} \) \(\begin{array}{l}y = \int (e^{3a.logx} + e^{3x.loga}) dx =... View Article