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