Question

\(\displaystyle\int e^{x}\left(\log x+\frac{1}{x^{2}}\right)dx\) is equal to
(consider \(\log_e x = \log x\))

• A. \(e^{x}\log_e x + c\) : where c is an arbitrary constant
• B. \(e^{x}\left(\log_e x+\frac{1}{x}\right)+c\): where c is an arbitrary constant
• C. \(e^{x}\left(\log_e x-\frac{1}{x}\right)+c\): where c is an arbitrary constant
• D. \(e^{x}\cdot\frac{1}{x}+c\) : where c is an arbitrary constant

Hint:

Write the integrand as \(e^{x}[f(x)+f^{\prime}(x)]\) with \(f(x)=\log x-\frac{1}{x}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The integrand looks like \(e^{x}\left[f(x)+f^{\prime}(x)\right]\). Such integrals have a neat result. But here the second term is \(\frac{1}{x^{2}}\), so we must first find which function \(f\) gives a derivative that produces this term.

Step 2: Key Formula or Approach:
We know \(\displaystyle\int e^{x}\left[f(x)+f^{\prime}(x)\right]dx = e^{x}f(x)+c\). Since the options are given, we can also differentiate each option and see which one returns the integrand.

Step 3: Try \(f(x)=\log x - \frac{1}{x}\).
Then \(f^{\prime}(x)=\frac{1}{x}+\frac{1}{x^{2}}\).
So \(f(x)+f^{\prime}(x)=\log x-\frac{1}{x}+\frac{1}{x}+\frac{1}{x^{2}}=\log x+\frac{1}{x^{2}}\).
This matches the integrand exactly.

Step 4: Write the integral.
\[ \int e^{x}\left(\log x+\frac{1}{x^{2}}\right)dx = e^{x}\left(\log x-\frac{1}{x}\right)+c \]

Step 5: Check the other options.
Option 1: \(\frac{d}{dx}\left(e^{x}\log x\right)=e^{x}\left(\log x+\frac{1}{x}\right)\), which has \(\frac{1}{x}\) and not \(\frac{1}{x^{2}}\). So it is wrong.
Option 2: \(\frac{d}{dx}\left[e^{x}\left(\log x+\frac{1}{x}\right)\right]=e^{x}\left(\log x+\frac{1}{x}+\frac{1}{x}-\frac{1}{x^{2}}\right)\), which is not the integrand. So it is wrong.
Option 4: \(\frac{d}{dx}\left(\frac{e^{x}}{x}\right)=e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\), which has no \(\log x\). So it is wrong.

Final Answer:
The integral equals \(e^{x}\left(\log x-\frac{1}{x}\right)+c\), which is option 3. \[ \boxed{e^{x}\left(\log x-\frac{1}{x}\right)+c} \]