Question

\( \int \frac{xe^x}{(1+x)^2 \, dx \) is equal to:}

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

Hint:

Whenever you see $e^x$ multiplied by a rational function, always check if you can split the fraction into the form $f(x) + f'(x)$.

Answer 1

The Correct Option is B

Concept: This integral can be solved using the property \( \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \). We need to manipulate the numerator to create a structure that fits this formula.

Step 1: Manipulate the integrand.
Rewrite the numerator \( x \) as \( (x+1) - 1 \): \[ \int \frac{((x+1) - 1)e^x}{(1+x)^2} \, dx = \int e^x \left[ \frac{x+1}{(x+1)^2} - \frac{1}{(x+1)^2} \right] \, dx \] \[ = \int e^x \left[ \frac{1}{x+1} - \frac{1}{(x+1)^2} \right] \, dx \]

Step 2: Identify \( f(x) \) and \( f'(x) \).
Let \( f(x) = \frac{1}{x+1} \). Then, by the power rule, \( f'(x) = \frac{d}{dx}(x+1)^{-1} = -1(x+1)^{-2} = -\frac{1}{(x+1)^2} \).

Step 3: Apply the integration property.
Since the expression is in the form \( \int e^x [f(x) + f'(x)] \, dx \): \[ \text{Integral} = e^x \left( \frac{1}{x+1} \right) + C = \frac{e^x}{x+1} + C \]