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प्रश्न
\[\int e^x \left( \log x + \frac{1}{x^2} \right) dx\]
बेरीज
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उत्तर
\[\text{ Let I }= \int\left( \log x + \frac{1}{x^2} \right) e^x dx\]
\[ = \int e^x \left( \log x + \frac{1}{x} - \frac{1}{x} + \frac{1}{x^2} \right)dx\]
\[ = \int e^x \left( \log x + \frac{1}{x} \right)dx + \int e^x \left( - \frac{1}{x} + \frac{1}{x^2} \right)dx\]
Let: t = ex log x
⇒ dt = `(e^xlogx + e^x/x)dx`
p = `-(e^x)/(x)`
⇒ dp = `(-(e^x)/(x) + (e^x)/(x^2))dx`
\[ \therefore I = \int dt + \int dp\]
\[ = t + p + C\]
\[ = e^x \log x + e^x \frac{- 1}{x} + C\]
\[ = e^x \left( \log x - \frac{1}{x} \right) + C\]
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