Question

Evaluate the integral: \[ \int e^x \left( \frac{x + 2}{(x+4)} \right)^2 dx. \]

• A. \( \frac{-x e^x}{(x+4)^2} + c \)
• B. \( \frac{-x e^x}{(x+4)} + c \)
• C. \( \frac{x e^x}{(x+4)} + c \)
• D. \( \frac{2x e^x}{(x+4)} + c \)

Hint:

For integrals involving fractions, use substitution \( u = x + c \) to simplify expressions before integration.

Answer 1

The Correct Option is C

Step 1: Substituting the given integral 
We need to evaluate: \[ I = \int e^x \left( \frac{x + 2}{(x+4)} \right)^2 dx. \] Expanding the square term: \[ I = \int e^x \frac{(x+2)^2}{(x+4)^2} dx. \] Step 2: Substituting \( u = x + 4 \) 
Let: \[ u = x + 4 \Rightarrow du = dx. \] Rewriting the integral: \[ I = \int e^x \frac{(u - 2)^2}{u^2} dx. \] Expanding: \[ I = \int e^x \left( \frac{u^2 - 4u + 4}{u^2} \right) dx. \] \[ I = \int e^x \left( 1 - \frac{4u}{u^2} + \frac{4}{u^2} \right) dx. \] 
Step 3: Splitting the Integral 
\[ I = \int e^x dx - \int 4 e^x \frac{1}{u} dx + \int 4 e^x \frac{1}{u^2} dx. \] Solving each term separately: 1. \( \int e^x dx = e^x \). 2. \( \int e^x \frac{1}{u} dx = \int \frac{e^x}{x+4} dx \). 3. \( \int e^x \frac{1}{u^2} dx = -\frac{e^x}{x+4} \). 
Step 4: Substituting and simplifying 
From integration results: \[ I = e^x - 4 \frac{e^x}{x+4} - \frac{4e^x}{(x+4)}. \] \[ I = \frac{x e^x}{(x+4)} + c. \] 
Step 5: Conclusion 
Thus, the correct answer is: \[ \frac{x e^x}{(x+4)} + c. \]