Question

Evaluate the integral: \[ I = \int \frac{x + 3}{(x + 4)^2} e^x \,dx \]

• A. \( e^x \frac{1}{x + 4} + C \)
• B. \( e^{-x} \frac{1}{x + 4} + C \)
• C. \( e^{-x} \frac{1}{x - 4} + C \)
• D. \( e^{2x} \frac{1}{x - 4} + C \)

Hint:

For integrals involving fractions with polynomials: - Use substitution to simplify the denominator. - Consider integration by parts if terms appear in the numerator. - Recognizing standard forms helps in quicker evaluation.

Answer 1

The Correct Option is A

Step 1: Use substitution. Let: \[ u = x + 4 \quad \Rightarrow \quad du = dx. \] Rewriting the given integral: \[ I = \int \frac{(u - 1)}{u^2} e^{u - 4} \,du. \] Expanding: \[ I = \int \left( \frac{u}{u^2} - \frac{1}{u^2} \right) e^{u-4} \,du. \] \[ I = \int \left( \frac{1}{u} - \frac{1}{u^2} \right) e^{u-4} \,du. \] Step 2: Solve by integration by parts. Using integration by parts for: \[ \int \frac{1}{u} e^{u-4} \, du. \] Let: \[ v = \frac{1}{u}, \quad dv = -\frac{du}{u^2}. \] \[ w' = e^{u-4}, \quad w = e^{u-4}. \] Using integration by parts: \[ I = \frac{e^{u-4}}{u} + C. \] Substituting back \( u = x+4 \): \[ I = \frac{e^x}{x+4} + C. \] Thus, the final result is: \[ I = e^x \frac{1}{x+4} + C. \]