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प्रश्न
\[\int e^x \left( \log x + \frac{1}{x} \right) dx\]
योग
उत्तर
\[\text{ Let I }= \int e^x \left( \log x + \frac{1}{x} \right)dx\]
\[\text{ Here}, f(x) = \log x\]
\[ \Rightarrow f'(x) = \frac{1}{x}\]
\[\text{ put }\ e^x f(x) = t\]
\[ \Rightarrow e^x \log x = t\]
\[\text{ Diff both sides w . r . t x}\]
\[ e^x \log x + e^x \frac{1}{x} = \frac{dt}{dx}\]
\[ \Rightarrow e^x \left( \log x + \frac{1}{x} \right)dx = dt\]
\[ \therefore \int e^x \left[ \log x + \frac{1}{x} \right]dx = \int dt\]
\[ \Rightarrow I = t + C\]
\[ = e^x \log x + C\]
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