Question

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

Answer 1

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

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

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

Let: t = e^x log x

⇒ dt = `(e^xlogx + e^x/x)dx`

p = `-(e^x)/(x)`

⇒ dp = `(-(e^x)/(x) + (e^x)/(x^2))dx`

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

\[ = t + p + C\]

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

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