Question

$\int e^{x}[\frac{1}{1+x}-\frac{1}{(1+x)^{2}}]dx=$ ________.

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

Hint:

Always look for $f(x) + f'(x)$ when you see $e^x$ in an integral.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
This integral is in a special form that allows for a direct application of a standard integration formula. The form is \(\int e^x [f(x) + f'(x)] dx\).
Step 2: Key Formula or Approach
The key integration formula is:
\[ \int e^x [f(x) + f'(x)] dx = e^x f(x) + C \] This formula can be proven using integration by parts. We need to identify \(f(x)\) and check if the other term in the parenthesis is its derivative, \(f'(x)\).
Step 3: Detailed Explanation
1. Identify \(f(x)\) and \(f'(x)\).
The integral is \(\int e^x \left(\frac{1}{1+x} + \left(-\frac{1}{(1+x)^2}\right)\right) dx\).
Let's try to set \(f(x) = \frac{1}{1+x}\).
Now, let's find the derivative of this function, \(f'(x)\).
\[ f(x) = (1+x)^{-1} \] Using the power rule and chain rule:
\[ f'(x) = -1 \cdot (1+x)^{-2} \cdot \frac{d}{dx}(1+x) \] \[ f'(x) = -1 \cdot (1+x)^{-2} \cdot 1 = -\frac{1}{(1+x)^2} \] 2. Check if the integrand matches the form.
The integrand is \(e^x \left(\frac{1}{1+x} - \frac{1}{(1+x)^2}\right)\).
We have identified \(f(x) = \frac{1}{1+x}\) and we found its derivative to be \(f'(x) = -\frac{1}{(1+x)^2}\).
So, the integrand is indeed in the form \(e^x[f(x) + f'(x)]\).
3. Apply the formula.
Using the formula \(\int e^x [f(x) + f'(x)] dx = e^x f(x) + C\):
\[ \int e^x \left(\frac{1}{1+x} - \frac{1}{(1+x)^2}\right) dx = e^x \cdot \frac{1}{1+x} + C \] \[ = \frac{e^x}{1+x} + C \] Step 4: Final Answer
The integral is \(\frac{e^x}{1+x} + C\).