Question

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

• A. $e^x \left\{ \log x - \frac{1}{x} \right\} + c$
• B. $e^x \left( \log x + \frac{1}{x} \right) + c$
• C. $e^x \{ \log x \} + c$
• D. None

Hint:

When an integrand with $e^x$ doesn't immediately show a function and its derivative, look for "hidden" pairs by adding and subtracting terms like $1/x$, $1/x^2$, or trigonometric functions.

Answer 1

The Correct Option is A

To solve the integral \( \int e^x \left( \log x + \frac{1}{x^2} \right) \, dx \), we will use the method of integration by parts and substitution.

The integration by parts formula is given by:

\(\int u \, dv = uv - \int v \, du\)

First, let's rewrite and split the integral as:

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

We will solve each part separately.

1. Part 1: \(\int e^x \log x \, dx\)
   • For this part, use integration by parts with \(u = \log x\) and \(dv = e^x \, dx\).
   • Then, we have \(du = \frac{1}{x} \, dx\) and \(v = e^x\).
   • Applying the integration by parts formula:
   • \(\int e^x \log x \, dx = e^x \log x - \int e^x \frac{1}{x} \, dx\)
2. Part 2: \(\int e^x \frac{1}{x^2} \, dx\)
   • Use substitution. Let \(v = \frac{1}{x}\), which gives \(dv = -\frac{1}{x^2} \, dx\).
   • Rewrite the integral as \(\int e^x \left(-dv\right)\), which simplifies to:
   • \(- \int e^x \, dv\)
   • Integrating directly, we get \(-e^x \frac{1}{x}\).

Combining the results from the two parts, we have:

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

Simplifying further, the integral becomes:

\(e^x \log x - \int e^x \frac{1}{x} \, dx - e^x \frac{1}{x} = e^x \log x - e^x \frac{1}{x}\)

Thus, the complete integrated result is:

\(e^x \left( \log x - \frac{1}{x} \right) + c\)

Therefore, the correct answer is: \(e^x \left\{ \log x - \frac{1}{x} \right\} + c\)