Question

The value of \[ \int e^{x} \left(\frac{x^{2} + 4x + 4}{(x+4)^{2}}\right)\, dx \] is:

• A. $e^{x}\left(\frac{x}{x+4}\right) + c$, where $c$ is a constant of integration.
• B. $e^{x}\left(\frac{4}{x+4}\right) + c$, where $c$ is a constant of integration.
• C. \(e^{x}\left(\frac{x}{(x+4)^{2}}\right) + C\)
• D. \(e^{x}\left(\frac{4}{(x+4)^{2}}\right) + C\)where $c$ is a constant of integration.

Hint:

When integrating expressions of the form $\int e^{x} g(x) dx$, always check if $g(x)$ can be broken down into $f(x) + f'(x)$. If it can, the integral simplifies directly to $e^{x}f(x) + c$.

Answer 1

The Correct Option is A

Step 1: Observe the integral: \[ \int e^{x} \left(\frac{x^{2}+4x+4}{(x+4)^{2}}\right) dx \] Rewrite the numerator: \[ x^{2}+4x+4 = x(x+4) + 4 \] So, \[ \frac{x^{2}+4x+4}{(x+4)^{2}} = \frac{x(x+4)+4}{(x+4)^{2}} \] 
Step 2: Split the expression: \[ \int e^{x} \left[\frac{x(x+4)}{(x+4)^{2}} + \frac{4}{(x+4)^{2}}\right] dx \] \[ = \int e^{x} \left[\frac{x}{x+4} + \frac{4}{(x+4)^{2}}\right] dx \] 
Step 3: Let \[ f(x) = \frac{x}{x+4} \] Differentiate using quotient rule: \[ f'(x) = \frac{(1)(x+4) - x(1)}{(x+4)^2} = \frac{4}{(x+4)^2} \] 
Step 4: Now the integral becomes: \[ \int e^x [f(x) + f'(x)] dx \] Using identity: \[ \int e^x [f(x) + f'(x)] dx = e^x f(x) + C \] \[ = e^x \left(\frac{x}{x+4}\right) + C \]  
Final Answer:
\[ \boxed{e^x \left(\frac{x}{x+4}\right) + C} \]