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: Understanding the Concept:
We need to integrate a product of an exponential function and a rational function. The structure of the terms in the parenthesis suggests using the special integral form involving \( e^x \) and a function plus its derivative.
Step 2: Key Formula or Approach:
The key formula is:
\[ \int e^x [f(x) + f'(x)] dx = e^x f(x) + C \] First, we can rewrite the integral by factoring out the constant \( e^1 = e \).
\[ \int e \cdot e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) dx = e \int e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) dx \] Now we need to check if the expression in the parenthesis is of the form \( f(x) + f'(x) \).
Step 3: Detailed Explanation:
Let's try to set \( f(x) \) to be the first term in the parenthesis.
Let \( f(x) = \frac{1}{x^2} = x^{-2} \).
Now, let's find its derivative, \( f'(x) \).
Using the power rule for differentiation:
\[ f'(x) = -2x^{-2-1} = -2x^{-3} = -\frac{2}{x^3} \] The expression in the parenthesis is exactly \( f(x) + f'(x) \).
So, the integral part fits the standard form:
\[ \int e^x \left( \frac{1}{x^2} - \frac{2}{x^3} \right) dx = e^x f(x) + C' = e^x \cdot \frac{1}{x^2} + C' \] Now, we include the constant 'e' that we factored out earlier:
\[ e \left( e^x \frac{1}{x^2} + C' \right) = \frac{e \cdot e^x}{x^2} + eC' \] Since \( e \cdot e^x = e^{x+1} \) and \( eC' \) is just another arbitrary constant C, the final result is:
\[ \frac{e^{x+1}}{x^2} + C \] Step 4: Final Answer:
The integral is \( \frac{e^{x+1}}{x^2} + C \).