Question

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

• A. $xe^{(x+\frac{1}{x})}+C$
• B. $e^{(x+\frac{1}{x})}+C$
• C. $x+e^{(x+\frac{1}{x})}+C$
• D. $x^{2}e^{(x+\frac{1}{x})}+C$
• E. $e^{(x+\frac{1}{x})}+x^{2}+C$

Hint:

Always check if the complex exponent's derivative is present as a multiplier.

Answer 1

The Correct Option is B

Step 1: Concept
Let $u = x + \frac{1}{x}$.

Step 2: Analysis
$du = (1 - \frac{1}{x^2}) dx = (\frac{x^2 - 1}{x^2}) dx$.

Step 3: Calculation
The integral becomes $\int e^u du = e^u + C$. Substitute back: $e^{(x + 1/x)} + C$. Final Answer: (B)