Question

\[\int e^x \left( \log x + \frac{1}{x^2} \right) dx\]

Answer 1

\[\text{ Let I }= \int\left( \log x + \frac{1}{x^2} \right) e^x dx\]

\[ = \int e^x \left( \log x + \frac{1}{x} - \frac{1}{x} + \frac{1}{x^2} \right)dx\]

\[ = \int e^x \left( \log x + \frac{1}{x} \right)dx + \int e^x \left( - \frac{1}{x} + \frac{1}{x^2} \right)dx\]

\[\begin{array} "let \text{  e}^x \log x = t \\ Diff\ both\ sides\ \\ \left( e^x \log x + e^x \frac{1}{x} \right)dx = dt \end{array}\begin{array} |"let  \text{ e}^x \left( - \frac{1}{x} \right) = p \\ Diff\ both\ sides\ \\ \left( e^x \frac{- 1}{x} + e^x \frac{1}{x^2} \right)dx = dp\end{array}\]

\[ \therefore I = \int dt + \int dp\]

\[ = t + p + C\]

\[ = e^x \log x + e^x \frac{- 1}{x} + C\]

\[ = e^x \left( \log x - \frac{1}{x} \right) + C\]