Question

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

• A. \(e^x \left( \log x + \frac{1}{x} \right) + C\)
• B. \(e^x \left( \log x - \frac{1}{x} \right) + C\)
• C. \(e^x \left( \log x + \frac{2}{x} \right) + C\)
• D. \(e^x \left( \log x - \frac{2}{x} \right) + C\)

Hint:

When dealing with expressions like \(e^x(\log x + \text{term})\), break it up and apply integration by parts where needed.

Answer 1

The Correct Option is B

Let: \[ I = \int e^x \left( \log x + \frac{1}{x^2} \right) dx \] Use integration by parts: Let \(u = \log x\), \(dv = e^x dx\). Then, \[ du = \frac{1}{x} dx,\quad v = e^x \] So, \[ \int e^x \log x \, dx = e^x \log x - \int \frac{e^x}{x} dx \] Now consider full integral: \[ I = \int e^x \log x \, dx + \int \frac{e^x}{x^2} dx \] Combining known results: \[ \int e^x \left( \log x + \frac{1}{x^2} \right) dx = e^x \left( \log x - \frac{1}{x} \right) + C \]